User:CogitoErgoCogitoSum/sandbox

This page associated to the Poncelet–Steiner theorem describes a number of Steiner constructions and related insights and alternative constructions, that are of unique interest for those fascinated by the Poncelet–Steiner theorem and its variants (such as portions of the arc, omission of circle centers, etc.), including the inline references to constructions with parallelograms, squares, etc., as well as other alternative constructions. This page exists to elaborate on the discussions of the Poncelet-Steiner theorem, and develop projective geometry tools to assist in the universality, generality, convenience, and insights of that theorem.

The constructions described below are ordered such that if another construction is required, it is prior listed. No construction requires another listed later, though some may reference constructions found in the Poncelet-Steiner theorem main article. There may also be passive references to other tertiary constructions reserved for yet another subpage. It is assumed that the reader/geometer has read and understood the main article, and has a grasp of the constructions outlined there.

The Projective Harmonic Conjugate
In this section we revisit and generalize two constructions found in the Poncelet-Steiner Theorem main article. The uses for the generalization will be justified in later sections, and the particular cases are demonstrated in this section.

Construction of the Projective Harmonic Conjugate
Given a line with segment $\overline{AB}$, and a point $C$ (or $D$) on the line, we may construct the projective harmonic conjugate, point $D$ (or $C$). The points $C$ and $D$ are each projective harmonic conjugates of one another with respect to, and defined by, the segment $\overline{AB}$. Indeed, it is also true that points $A$ and $B$ are each projective harmonic conjugates of one another with respect to the segment $\overline{CD}$, though in this case the latter point pair was determined from the former.

The projective harmonic conjugates define a proportionality with the segment $\overline{AB}$. It is the case that $$\frac{\overline{AC}}{\overline{AD}}=\frac{\overline{BC}}{\overline{BD}}$$.

The conjugates may be constructed thus:
 * 1) Choose a point $P$ arbitrarily in the plane.
 * 2) Draw lines $\overline{AP}$ and $\overline{BP}$ (in red).
 * 3) Choose a point $Y$ arbitrarily in the line $\overline{AP}$.
 * 4) Draw the line $\overline{BY}$ (in orange).
 * 5) Draw the line $\overline{CY}$ (in green):
 * 6) * This line intersects line $\overline{BP}$ at a point $X$.
 * 7) Draw the line $\overline{AX}$ (in magenta):
 * 8) * This line intersects line $\overline{BY}$ at a point $Q$.
 * 9) Draw the line $\overline{PQ}$ (in blue):
 * 10) * This line intersects line $\overline{AB}$ at a point $D$.
 * 11) Points $C$ and $D$ are conjugates of one another with respect to segment $\overline{AB}$.
 * 12) * It is also the case that points $A$ and $B$ are conjugates of one another with respect to segment $\overline{CD}$.

Alternatively, point $C$ may be constructed from point $D$ using the same set of lines, in which point $Q$ is found before point $X$.

Parallel from a bisected segment (revisited)
The parallel line construction from a bisected line-segment, as seen in the Poncelet-Steiner Theorem article, is a special case of the Projective Harmonic Conjugate construction. In this case, because one of the key defining points of the harmonic is a midpoint, point $C$, the conjugate point $D$ exists "at infinity" (and thus fails to exist on the line), thereby compelling the line $\overline{PQ}$ to be parallel. As such, there is no point $D$ on line $\overline{AB}$ that represents the points of conjugacy. In this construction, however, point $D$ is not the construction of interest, but rather its associated line, the (blue) line $\overline{PQ}$.

Midpoint from two parallels (revisited)
The construction of the midpoint from two parallels, as seen in the Poncelet-Steiner Theorem article, is a special case of the Projective Harmonic Conjugate construction. In this case, because one of the key defining points, $D$, of the harmonic does not exist (or rather exists "at infinity"), due to the two lines being parallel, its conjugate placed interior to the segment $\overline{AB}$, point $C$, must be the midpoint. The midpoint construction involves the same set of lines and points as the parallel line construction or the harmonic conjugate construction, simply in a different order and different set of initial conditions.

Construction of a Concurrent Line
Given any two lines, $m$ and $n$, a concurrent line through a point, $P$, is a line that intersects both lines $m$ and $n$ at the same point that lines $m$ and $n$ intersect one another. In other words, if lines $m$ and $n$ intersect at an unconstructed point $Z$ (off the page, not visible, at great distance), then the concurrent line is the line $$\overline{PZ}$$. Concurrent lines all intersect at the same point; from any two lines, concurrent lines may be constructed through any point in the plane.

The construction is convenient when a third line is required, defined by a point P and the (possibly unconstructed) intersection point between two other lines. This is a particularly useful construction when the point of intersection of the two lines is off the page at a great distance. The circle in the plane is not required.

Concurrent by way of projective points
Given lines $m$ and $n$ (both in black), and a point $P$ through which the concurrent is to be constructed:
 * 1) Draw two arbitrary lines through point $P$ (both in red), intersecting lines $m$ and $n$ at two points each:
 * 2) * One line through point $P$ intersects the lines $m$ and $n$ at points $A$ and $C$, respectively.
 * 3) * One line through point $P$ intersects the lines $m$ and $n$ at points $B$ and $D$, respectively.
 * 4) Draw two lines, $\overline{AD}$ and $\overline{BC}$ (both in green).
 * 5) * These green lines intersect at a point $X$.
 * 6) Choose an arbitrary point $E$ on line $m$ and draw the line $\overline{EX}$ (in orange).
 * 7) * Line $\overline{EX}$ intersects line $n$ at a point $F$.
 * 8) Draw lines $\overline{CF}$ and $\overline{EB}$ (both in magenta).
 * 9) * The lines intersect at a point $Q$.
 * 10) Draw the desired concurrent line $\overline{PQ}$ (in blue).

Special Case
It is important to note that whenever point $P$ is equidistant between the two lines $m$ and $n$ (i.e. existing on the angle bisector between them), the concurrent line construction will fail. Alternatively phrased, the construction will fail if the point $P$ is the midpoint of any segment colinear with the transversal line, in which consecutive interior angles are congruent, and the endpoints of the segment exist on the two lines $m$ and $n$. This is a special case scenario.

A concurrent through $P$ may still be constructed, however. A simple solution is to construct an intermediary concurrent line first.
 * 1) Choose an arbitrary point $S$ in the plane.
 * 2) * The point cannot be on either lines $m$ or $n$, nor on the concurrent through $P$.
 * 3) ** If the point $S$ is on the concurrent through $P$, the construction will again fail; however, line $\overline{PS}$ is therefore the desired concurrent.
 * 4) Construct a concurrent line to lines $m$ and $n$ through the point $S$.
 * 5) * This is possible since point $S$ is not equidistant between the two lines.
 * 6) From this new line and either one of the others, a concurrent through $P$ may now be constructed.
 * 7) * This is assured since point $P$ cannot be equidistant between the new line and either one of the other two.

Concurrent by way of the harmonic conjugate
Given any two lines $m$ and $n$, and an arbitrary point $P$ in the plane, the concurrent of the lines may be constructed through the point using the projective harmonic conjugate. This construction does not require a circle, and avoids the special case scenario of the projective point method.


 * 1) Draw an arbitrary line through point $P$ such that it cuts across lines $m$ and $n$.
 * 2) * Intersect these lines at points $A$ and $B$, respectively.
 * 3) Construct the projective harmonic conjugate of point $P$ with respect to segment $\overline{AB}$.
 * 4) * The conjugate is point $Q$, but it need not be constructed, and may exist "at infinity".
 * 5) * Any line passing through the conjugate, such as the one used to find $Q$ - a necessary part of the construction - even if it is parallel to line $\overline{AB}$, is the desired construction.
 * 6) * The goal is to have two lines, $\overline{AB}$ and one other, each of which pass through point $Q$, even if the point is not constructed.
 * 7) ** Let the second of these lines cross lines $m$ and $n$ at two points $C$ and $D$, respectively.
 * 8) Find the simple geometric center of the quadrilateral $ABDC$:
 * 9) * Intersect the lines connecting opposite vertices, assuming the quadrilateral is a simple shape.
 * 10) * These are lines $\overline{AD}$ and $\overline{BC}$.
 * 11) * The intersection is point $X$.
 * 12) Line $\overline{PX}$ is the desired concurrent.

Construction of an Additional Parallel From Any Two Parallels
Given two parallel lines, $m$ and $n$, a third parallel through an arbitrary point, $P$, can be constructed without the use of the circle.

This construction is merely a special case of the concurrent line construction, above, wherein the two given lines are already parallel. Thus, refer to the construction of the concurrent line in the previous section. It will not be repeated here. The parallel nature of lines $m$ and $n$ do not prohibit the concurrent line construction; the same set of constructable lines is achievable.

The principle behind this construction is that parallel lines "intersect at infinity", and concurrent lines intersect at the same point, therefore the concurrent to two parallels is also parallel.

Thus a third parallel through an arbitrary point may be constructed without a circle whenever there are already two parallels. The construction is trivial:
 * 1) Construct the concurrent through point $P$ of parallel lines $m$ and $n$.
 * 2) * This is the desired parallel through point $P$.

If the point $P$ exists equidistant between the two parallel lines, the same special case is applicable and the construction will fail, if the projective point method is used; use an intermediary parallel. If the harmonic conjugate method is used, this should not be an issue.

Constructing the Midpoint of a Line Segment
Given are three arbitrary lines in space - lines $a$, $b$, and $c$ - each parallel to one another, and the line in the middle is equidistant from the other two (i.e. "evenly spaced").



We may find the midpoint of any arbitrary line segment in space, $\overline{PQ}$:
 * 1) From an arbitrary endpoint - point $Q$ in this case - draw an arbitrary line (in red) cutting through the three parallels.
 * 2) * This line intersects the three parallels at points $J$, $K$, and $L$, respectively.
 * 3) Construct a parallel of the red line through point $P$ (in orange), the opposite endpoint of the line segment:
 * 4) * This is possible since the line segment $\overline{JL}$ is bisected by point $K$.
 * 5) * This line intersects the three parallels; one of the intersection points on the outer two lines is chosen: point $S$ on line $c$ in this case.
 * 6) * Observe that points $J$, $L$, and $S$ are three adjacent vertices of a parallelogram.
 * 7) Construct a line segment $\overline{JS}$ (in green), cutting through the middle of the three parallels.
 * 8) * This line segment - a diagonal of the parallelogram - intersects the middle of the three parallels at a point $T$.
 * 9) Construct the parallel of the red line through point $T$ (in blue).
 * 10) * This line intersects the line segment $\overline{PQ}$ at its midpoint $M$.

In an alternative construction, a midpoint of a line segment may be constructed from a parallelogram. The parallelogram allows for the construction of a line parallel to $\overline{PQ}$, upon which the midpoint construction found in the Poncelet-Steiner theorem main article may be utilized.

Three Evenly Spaced Parallels Imply a Parallelogram
Whenever three parallels of this type exist in the plane - with one line equidistant between the other two - any arbitrary line cutting through them can have a parallel of it constructed arbitrarily in space, as per steps 1 and 2 in the previous construction, because the three parallels intersect the cutting line such that a bisected segment exists. In the previous construction, points $J$, $L$, and $S$, form three adjacent of the four vertices of a parallelogram, the fourth being unlabeled. Thus, parallelograms are implied, and are arbitrarily constructable anywhere along the three parallels, such that any two of the parallels are opposing sides of the parallelogram. Any constructions involving the parallelogram are available.

Constructions with a Parallelogram
Parallelograms are a useful tool in their own right, though not as powerful as a circle.

Constructing a Second (Reflected) Parallel


Given a line $m$ and a parallelogram $STUV$ (both in black), a second parallel can be constructed symmetrically across the parallelogram:
 * 1) Draw two line segments $\overline{SU}$ and $\overline{TV}$ (both in green), connecting opposite points of the parallelogram.
 * 2) * These lines intersect at a point $X$, defining the center of the parallelogram.
 * 3) Extend each of the sides of the parallelogram by drawing lines $\overline{ST}$, $\overline{TU}$, $\overline{UV}$, and $\overline{SV}$ (each in brown).
 * 4) * These lines will always intersect line $m$ in at least two points.
 * 5) ** Just two intersections occurs when one set of sides of the parallelogram runs parallel to the given line $m$.
 * 6) ** In this case, two additional parallels to line $m$ are known to exist, and the construction may end here.
 * 7) * In the general case, as it is in this example, four points are generated: $A$, $B$, $C$, and $D$.
 * 8) Reflect any two of the points $A$, $B$, $C$, and $D$ across the parallelogram through point $X$:
 * 9) * Point $A$ is reflected by drawing line $\overline{AX}$ (in red), which intersects a side extension of the parallelogram at a point $A'$.
 * 10) Connect the reflected points with a line (in blue).

The principle is that parallels induce reflections. The parallelogram is not strictly speaking necessary; what is necessary is that there are two parallels intersecting the line being reflected, and an arbitrary point ($X$, in this case) that exists equidistant between the two parallels, across which we make reflections. The parallelogram allows for the construction of both the center point and the parallels, and because there are two distinct sets of parallels in a parallelogram we are assured the entire plane is covered.

Alternatively, line segment $\overline{SU}$ is bisected by point $X$. Two parallels may be made through points $T$ and $V$, as per the parallel line constructions found on the Poncelet-Steiner theorem page. These parallels will intersect the given line $m$, and together with point $X$ a reflection can be made.

Constructing Arbitrary Parallels When Given a Parallelogram
Regarding the above construction involving the parallelogram, the parallelogram predetermines the location of the second parallel, as a reflection across the parallelogram. However, once a second parallel is in the plane, a third parallel may be constructed arbitrarily. Whenever a parallelogram exists in the plane, any line may have a parallel constructed from it, through any arbitrary point in the plane.

Two previous constructions are trivially combined:
 * 1) From any line in the plane, a second (reflected, or symmetric) parallel may be constructed using the parallelogram.
 * 2) From two parallels, any third parallel may be constructed through any arbitrary point in the plane using the previous concurrent line constructions.

Parallelograms Imply Three Evenly Spaced Parallels
Given any parallelogram $ABCD$, three equally spaced parallels may be constructed. Indeed, four distinct sets may be constructed.


 * 1) Construct the center point $X$ of the parallelogram:
 * 2) * This is done by the intersections of the line segments connecting opposite vertices.
 * 3) * Point $X$ is equidistant between each pair of opposing sides, as well as each pair of opposite vertices.
 * 4) Any pair of opposite sides define parallel lines.
 * 5) * From any pair of parallels, a third parallel may be constructed through the center point $X$, using previous constructions.
 * 6) ** For example, we may construct lines $\overline{AB}$, $\overline{CD}$, and a third parallel through point $X$.
 * 7) * Two distinct sets of three equidistant parallels are definable in this way.
 * 8) Any diagonal connecting opposite vertices is bisected by the center point $X$.
 * 9) * From any bisected segment, a second (and a third) parallel may be constructed through the two available vertices.
 * 10) ** For example, we may construct line $\overline{BD}$ and the parallels through points $A$ and $C$.
 * 11) * Two distinct sets of three equidistant parallels are definable in this way.

Therefore, any constructions based on the existence of three equidistant parallels in the plane are achievable, whenever a parallelogram exists.

Since three equidistant parallels imply a parallelogram, and a parallelogram implies three equidistant parallels, they are equivalent premises in any construction.

Circles Imply Parallelograms


Given is a circle, centered at point $O$, we may construct an arbitrary inscribed parallelogram:
 * 1) Choose two points, $A$ and $B$, arbitrarily on the circle such that:
 * 2) * The circle center is not the midpoint.
 * 3) Draw the line segment $\overline{BC}$ (in red), denoting one side of the parallelogram.
 * 4) Draw lines $\overline{BO}$ and $\overline{CO}$ (in green).
 * 5) * These lines intersect the circle at points $A$ and $D$.
 * 6) Draw the line segments $\overline{CD}$, $\overline{DA}$, and $\overline{AB}$ (each in blue).
 * 7) Points $A$, $B$, $C$ and $D$ are the vertices of a parallelogram with known center $X$.
 * 8) * The parallelogram $ABDC$ is in fact a rectangle.

Any constructions involving the parallelogram may be achieved when a circle is provided.

The parallelogram does not imply the circle. The circle (with its center) is a stronger proposition.

Circles Imply Squares
Indeed, a circle implies a square, which is stronger than the parallelogram. Any line passing through the center of the circle may have a parallel of it constructed also through the circle. From the intersection points of these lines with the circle, an isosceles trapezoid (or triangle) is defined. The non-parallel sides of the trapezoid may be extended to an intersection point, which along with the circle center defines a perpendicular. The perpendicular intersects the circle at two points. The two perpendicular lines passing through the circles center define the diagonals of a square; the vertices are the points where these lines intersect the circle. Refer to the later construction Constructing a Centerline from Two Cutting Parallels for some details.

Translation
The translation of a line segment is a relatively trivial process, and can be done with either a circle or a parallelogram, or by any other means that allows for the construction of parallel lines.



Given is line segment $\overline{AB}$ (in black), and a point $A'$ to which the endpoint $A$ is to be translated.


 * 1) Draw a line $\overline{AA' }$ (in red).
 * 2) Construct a parallel of line $\overline{AA' }$ at point $B$ (in orange).
 * 3) Construct a parallel of line $\overline{AB}$ at point $A'$ (in magenta).
 * 4) Intersect these previous two parallel line constructions at a point $B'$.
 * 5) * Line segment $\overline{A'B' }$ (in blue) is the desired translation.

This construction will, however, fail if point $A'$ is colinear with line $\overline{AB}$. A simple solution for this condition is to translate the line segment $\overline{AB}$ to an intermediary non-colinear point $X$ first, then translate it back to point $A'$.

Rotation
The line segment rotation is another useful transformation, and is of particular use in Steiner constructions, as some constructions will fail under particular scenarios involving the coincidental orientation of line segments. The construction of the radical axis and the intersection between a line and a circle can suffer this scenario.

The Poncelet-Steiner theorem article covers one rotation method, which unfortunately fails when rotating a line segment a full 180 degrees. There is a workaround as explained in the article, but is unnecessarily burdensome. The construction provided below is an alternative approach to line segment rotation that is no more complicated than the one in the main article, and will work for half-circle rotations. This construction, however, has its own fallbacks.

Given is a line segment $\overline{AB}$ (in black), and an endpoint $A$ about which the rotation is made. This may represent a circle $A(B)$. Provided also is given circle $O(r)$.
 * 1) Construct the parallel of line $\overline{AB}$ (in red) through the given circle center, point $O$.
 * 2) * The parallel intersects the given circle at points $E$ and $F$.
 * 3) Choose one of the two points of intersection arbitrarily. Here point $F$ is chosen.
 * 4) Construct a line $\overline{AO}$ (in orange) through the points representing the centers of circles.
 * 5) Construct a line $\overline{BF}$ (also in orange) through the points representing the circumference of the circles.
 * 6) Intersect the orange lines at a perspective point $X$.
 * 7) * If point $X$ doesnt exist we may refer back to step 2 and choose the alternate point $E$ instead, and swap their roles for the remainder of this construction.
 * 8) * Alternatively if point $X$ doesnt exist or is at great distance, we may use concurrent line constructions between the orange lines wherever a line is to be constructed through point $X$.
 * 9) Let segment $\overline{OP}$ (in light green) represent an arbitrary line suggestive of the rotation to be made.
 * 10) * Point $P$ may be chosen arbitrarily or segment $\overline{OP}$ may be translated from elsewhere.
 * 11) * Point $P$ cannot lie on line $\overline{AO}$, or the construction will fail.
 * 12) * If point $P$ is chosen to be either points $E$ or $F$, whichever is not chosen in step 2, then a diametrically opposite, or half-circle rotation will be achieved.
 * 13) Construct a parallel of $\overline{OP}$ (in light blue) through point $A$, if it is not already provided.
 * 14) Construct a line $\overline{XP}$ (in dark blue), or the associated concurrent through point $P$.
 * 15) Intersect line $\overline{XP}$ with the light blue parallel of $\overline{OP}$.
 * 16) * This is point $B'$, the desired rotation.

As mentioned, if point $P$ is colinear with line $\overline{AO}$ this construction will fail. We may instead refer to the construction in the Poncelet-Steiner theorem article, or use a sequence of partial rotations to achieve the same ends, as was alternatively done in the main article.

Segment extensions
Given is a line segment $\overline{AB}$ (in black), on and extended by line $\overline{AB}$ (in red), and two arbitrary parallel lines $m$ and $n$ (both in orange) in the plane. One or both parallels may be constructed if necessary. We may construct on line $\overline{AB}$ an extension of the line segment to any arbitrary integer multiple. In the image below, segment $\overline{AC}$ is two times, segment $\overline{AD}$ is three times, and segment $\overline{AE}$ is four times the length of segment $\overline{AB}$.


 * 1) Choose a point $X$ arbitrarily on line $n$.
 * 2) * The point is a perspective point.
 * 3) Draw a line $\overline{AX}$ (in green).
 * 4) * The line intersects line $m$ at point $a$.
 * 5) Draw a line $\overline{BX}$ (in light blue).
 * 6) * The line intersects line $m$ at point $b$.
 * 7) Draw a line $\overline{aB}$ (in green).
 * 8) * The line intersects line $n$ at point $Y$.
 * 9) * The point $Y$ is a perspective point.
 * 10) Draw a line $\overline{Yb}$ (in light blue).
 * 11) * The line intersects line $\overline{AB}$ at point $C$.
 * 12) Point $C$ defines a segment extension by two.
 * 13) * Segment $\overline{AC}$ is twice the length of segment $\overline{AB}$
 * 14) * Segments $\overline{AB}$ and $\overline{BC}$ are congruent and colinear.
 * 15) Draw a line $\overline{CX}$ (in magenta).
 * 16) * The line intersects line $m$ at point $c$.
 * 17) Draw a line $\overline{Yc}$ (also in magenta).
 * 18) * The line intersects line $\overline{AB}$ at point $D$.
 * 19) * Point $D$ defines the extension by three times.
 * 20) Draw a line $\overline{DX}$ (in blue).
 * 21) * The line intersects line $m$ at point $d$.
 * 22) Draw a line $\overline{Yd}$ (also in blue).
 * 23) * The line intersects line $\overline{AB}$ at point $E$.
 * 24) * Point $E$ defines the extension by four times.
 * 25) This pattern of constructions (steps 9-10) may be continued indefinitely to arbitrary extension multiples.

Observe that on line $m$, a sequence of extensions are being generated, corresponding to those on line $\overline{AB}$. This second set of segment extensions are scaled down in size, proportional with the distances between the parallel lines. For the purposes of this construction, the proportionality, and thus the parallel line spacing is arbitrary. Specifically, the ratio between segments $\overline{AB}$ and $\overline{ab}$ is equal to the ratio of the distance line $\overline{AB}$ is to line $n$ and the distance line $m$ is to line $n$. This feature could be intentionally utilized if one has control over the line spacing.

The construction of the secondary segment extension, on line $m$, is essential for the following section on segment sectioning.

Extensions may also be achieved through a series of colinear translations of the segment, as seen in a previous section, which will require the construction of arbitrary parallels. This construction, though it makes use of parallels, they are not strictly arbitrary and are presumed to already be present.

Segment sectioning
Given is a segment $\overline{AB}$, which we wish to $n$-sect. In the example below, a trisection is performed. We are also given a parallel line $m$ on which there already exists a trisected segment $\overline{ad}$, which was constructed as a three-times extension of segment $\overline{ab}$, and as a by product of a three-times extension of segment $\overline{AB}$, as seen in the previous construction segment extensions.




 * 1) On parallel line $m$, construct an $n$-times extension of an arbitrary segment $\overline{ab}$.
 * 2) * In this example, $n=3$ and $\overline{ad}$ is the final extended segment, with points $b$ and $c$ trisecting it.
 * 3) Draw a line $\overline{Aa}$ (in magenta).
 * 4) * These points are chosen to connect one endpoint of the segment $\overline{AB}$ with one endpoint of the extended segment on line $m$.
 * 5) Draw a line $\overline{Bd}$ (also in magenta).
 * 6) * These points are chosen to connect the opposite endpoint of the segment $\overline{AB}$ with the opposite endpoint of the extended segment on line $m$.
 * 7) These lines intersect at a point $Z$.
 * 8) * Point $Z$ is a perspective point.
 * 9) * A point $Z$ can instead be found between the parallel lines, if lines $\overline{aB}$ and $\overline{Ad}$ had instead been chosen in the prior two steps.
 * 10) Through each point on line $m$ that extends segment $\overline{ab}$ and defines the $n$-sectioning on said line (namely points $b$ and $c$ in this example), construct a line through the perspective point $Z$ (each in blue).
 * 11) * These lines intersect segment $\overline{AB}$, thereby $n$-secting the segment.

Reflection of a point
Given is a line $m$ (in black) across which reflections are to be made, the point $P$ which we aim to reflect across the line, and the given circle (in green) centered at point $O$.

If line m passes through circle center O.

To reflect a single point about a line, the following construction is used:
 * 1) Construct the line $\overline{PO}$ (in red).
 * 2) * If point $P$ is coincident with point $O$ an alternative construction is required.
 * 3) Line $\overline{PO}$ intersects line $m$ at a point $Q$.
 * 4) * If line $\overline{PO}$ and $m$ are parallel, an alternative construction is required.
 * 5) Line $\overline{PO}$ intersects the given circle at a point $M$.
 * 6) Construct a line (in light green) which is parallel to line $m$, through point $M$.
 * 7) * This line intersects the given circle at a distinct point $N$.
 * 8) * If points $M$ and $N$ are coincident, an alternative construction is required.
 * 9) Draw a line $\overline{NO}$ (in pink).
 * 10) Construct a line (in purple) which is parallel to line $\overline{NO}$ through point $Q$.
 * 11) Construct through point $P$ a line (in blue) that is perpendicular to line $m$.
 * 12) * This line should intersect the previously constructed purple line at a point $P'$, the reflection of $P$ across the line $m$

Reflection of figures
To reflect any figure about a line - be they other lines, circles, or other figures - simply reflect each of its defining points about the line.

The construction is a trivial application of the point reflection:
 * 1) Identify the line $m$ across which the figure is to be reflected.
 * 2) Identify each of the $n$ key points ($P1$ to $Pn$) that uniquely define the figure.
 * 3) Perform a reflection for each of these $n$ points.

Construction of a Perpendicular from a Square
Given is a square $ABDC$ (in dark green) and an arbitrary line $m$ (in black), and a point in the plane $P$. We wish to construct a perpendicular of line $m$ through point $P$.




 * 1) Arbitrarily choose and draw a line through the diagonal of the square.
 * 2) * In this example, line $\overline{BC}$ (in gray) is drawn.
 * 3) From point $P$, two lines perpendicular to one another are drawn (in dashed light blue and dashed red), each parallel to one of the two adjacent sides of the square.
 * 4) * Since opposite sides of the square are parallel, we may use previous construction techniques.
 * 5) * These lines will intersect the gray line $\overline{BC}$ through the diagonal of the square at points $X$ and $S$, respectively.
 * 6) From points $X$ and $S$ new lines (in dashed green and dashed orange) are constructed parallel to the alternate sides of the square.
 * 7) * These lines are thus perpendicular to the previous set of lines constructed.
 * 8) * These lines intersect the line $m$ at points $Y$ and $T$, respectively.
 * 9) Like before, two new lines are drawn from points $Y$ and $T$ (in dashed light blue and in dashed red), parallel to the alternate square sides.
 * 10) * These lines will, respectively, intersect the gray diagonal at points $Z$ and $U$.
 * 11) Finally, the process is repeated one final time; two new lines parallel to the sides of the square are drawn from points $Z$ and $U$ (in dashed green and dashed orange).
 * 12) * These lines intersect one another at point $M$.
 * 13) The line $\overline{PM}$ (in dark blue) is constructed.
 * 14) * This is the desired line, perpendicular to line $m$.

Construction of a Tangent/Polar to a Circle or Arc
These constructions concern the construction of tangent lines, points of tangency, pole point, and the polar line, which runs through the points of tangency.

From a point on the arc of a circle
Given is an arbitrary circle (in green) with a center point $O$ (which will not be utilized), and a point $P$ existing on the arc of the circle. We wish to construct the tangent lines to the circle at the point. Such a point has a tangent line that is also the polar.


 * 1) We first arbitrarily pick four new points on the arc of the circle: $A$, $B$, $C$, and $D$, respectively.
 * 2) * These points are labeled sequentially and clockwise from point $P$.
 * 3) * Tangents may be drawn from any of these points.
 * 4) * Indeed, the arc of the circle need not be provided at all; it is sufficient to have the five points on the circumference defined.
 * 5) Construct an irregular pentagram (in a dark blue) between these five points on the arc of the circle.
 * 6) * Draw line segments $\overline{PB}$, $\overline{BD}$, $\overline{AD}$, $\overline{AC}$, and $\overline{PC}$.
 * 7) Draw line $\overline{AD}$ (in red).
 * 8) * The line is coincident with the edge of the pentagram closest to but not intersecting point $P$.
 * 9) * In other words, the pentagram line segments projecting from point $P$, when traced out, will cross the same line segment first.
 * 10) Find the intersection points $R$ and $S$.
 * 11) * Point $R$ is the point of intersection between segments $\overline{PB}$ and $\overline{AC}$.
 * 12) * Point $S$ is the point of intersection between segments $\overline{PC}$ and $\overline{BD}$.
 * 13) * Or in other words, the pentagram line segments projecting from point $P$, when traced out, will have a second crossing of a pentagram line segment.
 * 14) Draw the line $\overline{RS}$ (in orange).
 * 15) Lines $\overline{AD}$ and $\overline{RS}$ intersect at a point $X$.
 * 16) Line $\overline{PX}$ (in blue) is the desired tangent line, which is its own polar, to the circle at point $P$, which is its own inverse.

From a point not on the circle
There are numerous properties involving the polar of which we may take advantage. Several different constructions avail themselves, and are useful under different circumstances.

These constructions work whether point $P$ is internal or external to the circle, but will fail if it is on the arc of the circle or coincident with the circle center. In the former case, refer to the previous construction for a tangent line through a point on the circle. In the latter case, no polar exists.

Using three cutting lines through the circle
Given is an arbitrary circle (in green) with a center point $O$ (which will not be utilized), and a point $P$ existing not on the circle. We wish to construct the tangent lines to the circle which pass through the point. Observe the relationships between the points of tangency on the circle, tangent lines through point $P$, the polar line of point $P$, and the inverse in the circle of point $P$.


 * 1) We first construct three arbitrary lines (in gray) passing through point $P$, cutting through the circle and intersecting it in at least one point for each line.
 * 2) * One line intersects the circle at points $A$ and $B$.
 * 3) * One line intersects the circle at points $C$ and $D$.
 * 4) * One line intersects the circle at points $E$ and $F$.
 * 5) * If one of the arbitrary lines has one intersection point (e.g. point $A=B$) then this is a tangent line, and the point (e.g. $A$) is the point of tangency, existing on the polar.
 * 6) Draw line segments $\overline{AD}$, $\overline{BC}$, $\overline{CF}$, and $\overline{DE}$ (each in red).
 * 7) * These line segments will intersect at points $X$ and $Y$.
 * 8) * If $A=B$ then $X=A=B$.
 * 9) Draw a line $\overline{XY}$ (in blue).
 * 10) * This line is known as a polar of point $P$ to the circle.
 * 11) Line $\overline{XY}$ intersects the circle at points $S$ and $T$.
 * 12) * Points $S$ and $T$ are the points of tangency.
 * 13) Construct lines $\overline{PS}$ and $\overline{PT}$.
 * 14) * These lines are the tangent lines to the circle through point $P$.

Using two cutting lines through the circle
It is valid to point out that, strictly speaking, points $E$ and $F$ and the line passing through them never needed to be drawn. Points $A$, $B$, $C$ and $D$ were sufficient. The point $Y$ may be constructed as the intersection of lines $\overline{AC}$ and $\overline{BD}$, although this is often less convenient. Additionally, point $Y$ need not be constructed; we need only the concurrent between lines $\overline{AC}$ and $\overline{BD}$ passing through point $X$; this is the polar in the circle of point $P$.

Alternatively, we could have used points $A$ and $B$ (which exist on the same line through point $P$). From these points we could have constructed tangents to the circle, as per the previous construction for a tangent through a point on the arc of the circle. These tangent lines would intersect one another on the polar at a point $X$. We can then find point $Y$, also on the polar, by finding the intersection of the tangents to the circle at points $C$ and $D$ (both of which are colinear with point $P$). As before, concurrent line constructions may be employed when necessary.

Alternative involving the Harmonic Conjugate
We may also find a point $X$ as the harmonic conjugate of point $P$ on the line $\overline{AB}$. Point $Y$ may be found as the harmonic conjugate of point $P$ on the line $\overline{CD}$. The projective harmonic conjugate of point $P$ on any cutting line through the circle exists on the polar. It should be noted that any chord through the circle on the line passing through point $P$ defines a line segment upon which the projective harmonic conjugate of the point lies on the polar. Using multiple such chords, multiple points on the polar may be constructed, thus defining the line.

Construction of the Inverse in a Circle of a Point on the Centerline
This is a relatively trivial construction. Let point $P$ exist in the plane other than the center of the circle, point $O$. For point $O$ no inverse exists.

For any circle with radius $r$ and center point $O$, any point $P$ in the plane and its inverse (point $I$) in the circle share a proportionality relationship. It is the case that $$\overline{PO}\cdot\overline{IO}=r^2$$.

Let also a line exist through point $P$ that is centerline through the circle (that is, it passes through the center of the circle). The center point of the circle need not be constructed or provided. Naturally if the centerline does not exist, it must be constructed, either directly as the line through the circle center or through some other means.

The inverse of point $P$ may be constructed simply.

If point $P$ exists on the arc of the circle, nothing need be done. The point is its own inverse.

Let point $P$ exist in the plane internal or external to the circle:
 * 1) Construct the polar to the circle of point $P$.
 * 2) Where the polar intersects the centerline, the inverse is constructed.

Alternatively, if the point $P$ is internal the circle, a line through it perpendicular to the centerline may be constructed. Where this perpendicular intersects the circle, tangents to the circle may be constructed, which intersect one another at the inverse point. The line perpendicular to the centerline through point $P$ may be constructed as the third parallel to the two tangents to the circle at the points the centerline intersect the circle.

Given the polar construction of the previous section using the projective harmonic conjugate technique, if the chord is chosen to be the diameter of the circle, the constructed conjugate is also the inverse of a point $P$ in a circle.

Constructing the Pole from a Polar in a Circle
Given an arbitrary line $m$ in space and a circle (devoid of its center), we may find the point $M$ that is the pole to the polar $m$ in the circle.

This construction will work for polar lines that are both intersecting and non-intersecting to the circle.


 * 1) Choose any two points $A$ and $B$ arbitrarily on the line $m$.
 * 2) To the circle, construct the polar of the point $A$.
 * 3) * The polar is line $a$.
 * 4) To the circle, construct the polar of the point $B$.
 * 5) * The polar is line $b$.
 * 6) Intersect the lines $a$ and $b$.
 * 7) * The intersection point is $M$, the pole associated with the polar line $m$.

This construction is fully general. Each of the special cases to follow are simplified special cases of this construction.

from a tangential polar
This is a trivial case. If the polar is tangential to the circle, the point of intersection to the circle is the pole. The pole is on the polar at the point of tangency. There is nothing to construct, but to find the point of tangency by intersecting the circle with the tangent.

from a polar cutting through the circle
For any polar intersecting the circle in two points $A$ and $B$, we may construct the tangents to the circles at the points $A$ and $B$. These tangent lines intersect one another at a unique point exterior to the circle. The intersection point is the pole.

If the polar is also a centerline (thus crossing the circle center) then there is no pole; or it "exists at infinity". The tangent lines to the circle, in this case, will be parallel.

Constructing a Centerline from Two Parallels
Given are two parallel lines and a circle placed arbitrarily in the plane. Construct additional parallel lines as necessary, such that no fewer than two intersect the circle, with a total of four distinct intersection points between them.

Let one of the parallel lines intersect the circle at points $A$ and $B$; let a second parallel intersect at points $C$ and $D$. A centerline to the circle may be constructed simply:
 * 1) Bisect line segment $\overline{AB}$.
 * 2) * Let the midpoint be point $M$.
 * 3) * The bisection may be done using the segment bisection construction by way of two parallels, as found on the Poncelet-Steiner theorem main article.
 * 4) Similarly, bisect line segment $\overline{CD}$.
 * 5) * Let the midpoint be point $N$.
 * 6) Construct the line $\overline{MN}$.
 * 7) * This line is perpendicular to the two parallel lines.
 * 8) * This line is a centerline to the circle.

Observe that a single midpoint construction (or segment bisection) may be used. Line segments $\overline{AB}$ and $\overline{CD}$ may be bisected concurrently, and bisected by the very centerline we wish to construct. The details may be found in the appropriate section on the Poncelet-Steiner theorem main article.

More concisely, we may define points $X$ and $Y$ by the intersections of lines $\overline{AC}$ with $\overline{BD}$ and $\overline{AD}$ with $\overline{BC}$, respectively. The line $\overline{XY}$ is the desired centerline (which bisects both $\overline{AB}$ and $\overline{CD}$ concurrently).

Should a point (e.g. $X$) not exist due to two lines (e.g. $\overline{AC}$ with $\overline{BD}$) being parallel, we may use concurrent line constructions, or we may use the two parallel lines $\overline{AB}$ and $\overline{CD}$ to construct a third parallel through the circle at a more convenient location for this construction. Alternatively, we may recognize that this scenario can only happen if $ABCD$ is a rectangle, in which case each diagonal defines a centerline.

If one of the parallels happens to pass through the circle center, its intersection points with the circle defines two opposing vertices of a square inscribed in the circle. The constructed centerline, which is orthogonal and also passes through the circle center, defines the other two vertices. Thus a square may be defined, as per a previous section Circles Imply Squares.

alternative with poles
Alternatively, we may find the two distinct poles corresponding to the two distinct parallel lines (treated as polars) to the circle. The two poles of these parallels define a perpendicular line which passes through the circle center.

Constructing a common centerline from a point on the centerline

 * Generalize this section to point that is inside one or two circles.
 * Write in terms of centerline from two parallels, in previous section.



Given two non-intersecting circles (both in dark green), each devoid of their centers, and given an arbitrary point on the centerline, point $M$, exterior to at least one of the two circles, it is possible to construct the centerline.


 * 1) From the point $M$ on the centerline, construct the points of tangency to each of the two circles.
 * 2) * The polar, line $\overline{ST}$ (in red), connect the points of tangency of the near circle.
 * 3) * The polar, line $\overline{UV}$ (in orange), connect the points of tangency of the far circle.
 * 4) * If a polar of one of the circles cannot be constructed then see the notes below for an alternative construction.
 * 5) Draw lines from each point of tangency on the near circle, to a point of tangency on the other circle:
 * 6) * Construct line $\overline{SV}$ (in light green).
 * 7) * Construct line $\overline{TU}$ (in light blue).
 * 8) * The points of tangency should be paired consistently. Lines $\overline{SU}$ and $\overline{TV}$ also work.
 * 9) The lines intersect at a point $X$, which is also on the centerline.
 * 10) Construct line $\overline{MX}$ (in dark blue).
 * 11) * This is the centerline.

If the point $M$ is in the interior of one of the circles, only a slight modification is necessary. The polar through only one of the circles is truly necessary. Tangent lines through the point $M$ to any circle they can be constructed on will intersect the other circle at two points for each tangent line; they will be symmetric across the centerline. In the general case at least two additional centerline points may be constructed as intersections.

This construction will fail if the point $M$ is interior to both circles simultaneously, one interior to the other but not intersecting. It will also fail if the point is the common point of tangency between the two circles, be they interior or exterior.

In general, from any point on the centerline, two parallel lines may be constructed which are perpendicular to the centerline. From this scenario, refer back to the centerline construction from two parallel lines in the plane.

Constructing a common centerline from a point on the radical axis
When two arbitrary non-intersecting circles are provided - left and right circles $l$ and $r$ - without their centers, and a point $P$ exists in the plain on the radical axis between the circles, the common centerline through the circles may be constructed.




 * 1) From point $P$, construct the tangent points to circle $l$: points $A$ and $B$.
 * 2) * The tangent lines are drawn for convenience (in dashed black), but are themselves unimportant for construction. Only the points of tangency are significant.
 * 3) From point $P$, construct the tangent points to circle $r$: points $C$ and $D$.
 * 4) * The tangent lines are drawn for convenience (in dashed black).
 * 5) From each of the two points of tangency on either circle, draw lines to the points of tangency on the other circle.
 * 6) * Lines $\overline{AC}$ and $\overline{BD}$ are both in light green.
 * 7) * Lines $\overline{AD}$ and $\overline{BC}$ are both in red.
 * 8) * There will always be a point of intersection, point $M$, between the two circles made by the red lines.
 * 9) * The green lines, however, if they intersect will do so to one side, away from and outside of the two circles, possibly at a great distance (point $X$). This happens if they are not congruent circles. If the circles are congruent then the lines will be parallel, and we may say point $X$ is "at infinity".
 * 10) Draw the line passing through point $M$ (in blue) that is concurrent with the green lines.
 * 11) * Draw the line $\overline{MX}$ if point $X$ exists.
 * 12) * Or construct the concurrent or, if necessary, a third parallel to the green lines, through the point $M$.
 * 13) The blue line $\overline{MX}$ is the centerline of the two circles, passing through the centers of each circle.

Construction of the Radical Axis From a Point on the Axis
This is a relatively trivial construction. The intersection point of the polars to each circle from the point on the radical axis, is also on the radical axis, and thus define the line:


 * 1) From the point $P$ on the radical axis, construct the polar to each circle.
 * 2) Intersect the polars at a point $X$, if they intersect.
 * 3) * If the point $P$ is also on the centerline between circles then the polars will be parallel.
 * 4) Draw or construct the line through the point $P$ and point $X$, if it exists.
 * 5) * One may construct the concurrent, or third parallel if necessary, of the polars of point $P$, through point $P$.
 * 6) * This is the radical axis.

Intersecting a Line with the Circle of an Arc
Given is an arc of a circle (in dark green), of arbitrary arc length, spanning between the endpoints $A$ and $B$ of the arc $\overarc{AB}$. Given also is a line $m$ (in black). We wish to intersect the line with the circle of the arc; that is to say, we wish to find the intersection points of the line with the circle which is defined by the arc.

Notice that for this construction the center, point $O$, of said circle is not utilized. It need not be provided at all in order to intersect a line with the arc of the circle. However, according to Steiner's theorem, in order to construct all of Euclid the center is still required, if provided only a single circular arc in the plane.




 * 1) At points $A$ and $B$, construct the tangent lines to the circle (in dashed gray).
 * 2) * These tangent lines intersect at a point $C$.
 * 3) * If these lines do not intersect then the tangents are parallel:
 * 4) ** Points $A$ and $B$ exist diametrically opposite the circle center.
 * 5) ** We treat the point $C$ as existing "at infinity"; any line passing through it may be constructed as another parallel.
 * 6) ** Alternatively, see notes below for a different treatment.
 * 7) Draw a line $\overline{AB}$ (in red).
 * 8) Lines $\overline{AB}$ and $m$ intersect at a point $D$.
 * 9) * If these lines do not intersect then they are parallel:
 * 10) ** We treat the point $D$ as existing "at infinity"; any line passing through it may be constructed as another parallel.
 * 11) ** Alternatively, see notes below for a different treatment.
 * 12) ** If the tangents to the arc were also parallel in step 1 then the alternative treatment is required. The tangents through points $A$ and $B$ cannot be parallel to one another and perpendicular to both lines $\overline{AB}$ and $m$, simultaneously.
 * 13) Draw a line $\overline{CD}$ (in orange).
 * 14) Choose two points arbitrarily on line $\overline{AB}$, named points $E$ and $F$.
 * 15) * Neither point can be point $D$, but is otherwise completely arbitrary.
 * 16) * Points $A$ and $B$ may be chosen to coincide with $E$ and $F$, if convenient.
 * 17) Draw lines $\overline{CE}$ and $\overline{CF}$ (both in light green).
 * 18) * These lines intersect line $m$ at points $I$ and $G$, respectively.
 * 19) Draw a line $\overline{EG}$ (in pink).
 * 20) * This line intersects line $\overline{CD}$ at point $H$.
 * 21) Draw a line $\overline{HI}$ (in light blue).
 * 22) * This line intersects line $\overline{AB}$ at point $J$.
 * 23) Draw a line $\overline{CJ}$ (in purple).
 * 24) * This line intersects line $\overline{EG}$ at point $K$.
 * 25) Draw a line $\overline{DK}$ (in brown).
 * 26) * This line intersects the arc at two points $L$ and $M$.
 * 27) * If there is no intersection with the arc, then line $m$ does not intersect the circle of the arc.
 * 28) Draw lines $\overline{CL}$ and $\overline{CM}$ (both in dark blue).
 * 29) * These lines intersect line $m$ at points $X$ and $Y$, respectively.
 * 30) * These are the points of intersection between the line $m$ and the circle of the arc.

Note: In the event that lines are inconveniently parallel, we may resolve the dilemma by reducing the arc. We may choose a point arbitrarily on the arc, e.g. point $B`$, that truncates it into a smaller arc $\overarc{AB`}$. In doing this we may ensure that the tangent lines are not parallel, and/or that the lines $\overline{AB`}$ and $m$ are not parallel.

With this construction, the Poncelet-Steiner theorem can be strengthened, as per Francesco Severi's 1904 theorem.
 * Any compass and straightedge construction may be completed with the straightedge alone, provided that any portion of the arc of a single circle, along with the center of the circle of said arc, exists in the plane.

Circle Center from Two Centerlines (Basic Principle)


The center point of a circle is trivially determined from the centerline constructions, which are listed below, and depend on the scenario. The concept for constructing the center point of a circle is fairly straightforward:


 * 1) Construct one centerline arbitrarily on one of the circles.
 * 2) Construct a second, arbitrary but non-coincident, centerline on the same circle.
 * 3) The two centerlines intersect at a unique point, the center of the prescribed circle.

How these centerlines are constructed will depend on the information provided, and will be the principle of center constructions in the variety of arrangements to follow.

Construction of a Circle Center from Concentric or Intersecting Circles
Two circles without their centers, by itself, is not sufficient information to construct a center. Something more must be given. There are a number of ways to construct the center of a circle given two or more circles, and having additional but sufficient information about the scenario that allows the center to be recovered. This is not an exhaustive list.

Circle Center / Centerlines From Two Concentric Circles
Two circles without their centers is not sufficient information to construct a center. Something more must be given, and knowing the fact that they are concentric turns out to be sufficient information. If given two concentric circles devoid of a center, the center may in fact be constructed using a straightedge only.

Given below are two concentric circles (both in green): the inner circle, $i$, and the outer circle, $o$.




 * 1) Choose a point $A$ arbitrarily on the outer circle $o$.
 * 2) Using point $A$ as an external point to the inner circle $i$, construct the tangent lines (both in deep dark blue) to the inner circle.
 * 3) * The points of tangency are points $S$ and $T$.
 * 4) * The tangent lines will intersect the outer circle at points $B$ and $C$.
 * 5) From each points $B$ and $C$, construct tangent lines to the inner circle (both in blue).
 * 6) * Each point already has one of its two tangent lines constructed.
 * 7) * The points of tangency are points $U$ and $V$.
 * 8) Take note that this process may continue for as long as is preferred and convenient, constructing new tangent lines encircling the inner circle.
 * 9) Lines $\overline{AT}$ and $\overline{BU}$ intersect at a point $X$
 * 10) Lines $\overline{AS}$ and $\overline{CV}$ intersect at a point $Y$.
 * 11) Lines $\overline{BU}$ and $\overline{CV}$ intersect at a point $Z$.
 * 12) Observe the isosceles triangles and the deltoid shapes formed by these constructions (highlighted in shades of blue). These shapes are symmetric about their centerlines.
 * 13) * $ACY$ and $ABX$ form isosceles triangles. $ABZC$ forms a deltoid.
 * 14) * The two red lines are lines of symmetry across the two isosceles triangles, connecting a peak vertex of the triangle with a point of tangency of at the midpoint of the opposite side.
 * 15) * The orange line is the line of symmetry across a deltoid, connecting vertices.
 * 16) * These lines are all defined by the intersection points of tangent lines, and/or their points of tangency.
 * 17) * All three of these lines are centerlines to the circles, thus intersecting at the same point $O$, the center of the circles.

Circle Centerlines From Two Intersecting Circles
Two circles without their centers is not sufficient information to construct a center. Something more must be given, and intersection points is sufficient. When two circles are given devoid of their centers, if they intersect we may construct their centers using a Steiner construction.

There are two cases: When the two circles intersect at two points, and alternatively when they intersect at one point (i.e. are tangent to one another). These must be treated separately.

Centers are determined from the intersections of two or more centerlines. To find a circle center in either of the two cases, each of the constructions below must be completed twice, using unique points where arbitrary points are required.

Centerline from two unique points of intersection

 * Show the similarity to the tangential circle construction.

Below two intersecting circles are given (in dark green), devoid of centers, and intersecting at two points $U$ and $V$.




 * 1) The two circles intersect at two distinct points $U$ and $V$.
 * 2) Choose a circle arbitrarily in which the centerline is to be constructed. In this example the larger circle on the left is chosen.
 * 3) Using the alternate circle - the smaller one on the right - arbitrarily choose two initial points $A$ and $B$.
 * 4) Construct four lines (in dark blue), each one passing through either one point $A$ or $B$, and passing through either one point $U$ or $V$.
 * 5) * The lines associated with point $U$ intersect the larger circle at points $C$ and $D$.
 * 6) * The lines associated with point $V$ intersect the larger circle at points $E$ and $F$.
 * 7) Two new lines are drawn (in red), $\overline{CE}$ and $\overline{DF}$.
 * 8) * These lines cross one another in the circle interior.
 * 9) * These lines are associated with the points on the larger circle which are themselves associated with the same same arbitrary point on the smaller circle.
 * 10) * The two red lines intersect at a point $Y$.
 * 11) Two new lines are drawn (in orange), $\overline{CD}$ and $\overline{EF}$.
 * 12) * These lines intersect at a point $X$ exterior to the circle.
 * 13) The line $\overline{XY}$ (in light green) cuts the circle in half, symmetrically down its centerline, bisecting the chosen circle into two equal halves.

Centerline from one point of intersection (tangential circles)
Below two intersecting circles are given (in dark green), devoid of centers, and intersecting at one point of tangency, the point $I$.




 * 1) Place two points, $A$ and $B$, arbitrarily on either circle.
 * 2) Construct two lines through point $I$, the point of tangency, through each points $A$ and $B$ (in dark blue).
 * 3) * Lines $\overline{AI}$ and $\overline{BI}$ intersect the other circle at points $D$ and $C$, respectively.
 * 4) Draw lines $\overline{AB}$ and $\overline{CD}$ (each in red).
 * 5) * These two lines are parallel to one another.
 * 6) Choose which of the two circles is to have a centerline constructed through it. In this example the left, larger circle is chosen.
 * 7) Arbitrarily pick two points, $S$ and $T$, on the circle through which a centerline is to be made.
 * 8) Construct a line through each of the points $S$ and $T$ (in orange), parallel to the red lines $\overline{AB}$ and $\overline{CD}$.
 * 9) * This is possible as there are two parallels in the plane. A third parallel may be constructed from them through any arbitrary point in the plane.
 * 10) * These lines intersect the circle at points $U$ and $V$, respectively.
 * 11) Construct line segments $\overline{AU}$, $\overline{BS}$, $\overline{SV}$, and $\overline{UT}$ (in pink).
 * 12) * These line segments will intersect one another in the interior of the circle at points $X$ and $Y$.
 * 13) Construct line $\overline{XY}$ (in light green).
 * 14) * This is the desired centerline (diameter, line of symmetry), through the left, larger circle.

From Two Parallel Lines in the Plane
From any two parallel lines, additional parallels may be constructed. Construct zero-to-four additional parallels, as necessary, such that at least two parallel lines intersect each circle. From two parallel cutting lines through a circle, a centerline through each circle may be constructed perpendicular to these parallel lines. This technique is explained in the centerline from two cutting parallels and circles imply squares sections.

In the general case, there will be two distinct such centerlines, one for each circle. These constructed centerlines are parallel to one another, of which there are two; furthermore, they intersect the original two parallel lines, to which they are perpendicular, thus defining four points of intersection that encompass a rectangle (a parallelogram). From this information it is a triviality to find a second distinct centerline through each circle, using previously described techniques. One may take advantage of the two parallel centerlines to construct a second cutting parallel through each circle, or one may take advantage of the parallelogram to construct parallels of arbitrary lines. Using either approach, two cutting parallels through each circle defines a second distinct centerline.

If from the two original parallel lines only one common centerline through the circles is defined, the construction will fail. It happens by coincidence that the two parallel lines were perpendicular to a common centerline. This is a special case scenario. Constructing the common centerline from two parallels is a similar construction to that from a point on the centerline. An alternative construction exists to find the center points of two circles from the common centerline through them. This construction is explained in the next section.

From Two Circles With Both a Common Centerline and a Common Tangent
The common centerline, line $c$, already counts as a centerline through each circle, and thus one of the two centerlines required to find a center. We need only construct a second centerline through either one or both of the other two circles, thereby allowing a center point to be constructed. The common tangent line, line $t$, allows us to construct what is needed quite trivially. The center points may be constructed by way of two parallel lines in the plane, as per the previous construction discussed in this section. The parallel lines, however, are themselves trivially constructed.

Let points $A$, $B$, $C$, and $D$ be the four points of intersection between the common centerline, line $c$, and the two circles. Labeled the points in that same order, such that point $A$ on circle $l$ corresponds to point $C$ on circle $r$ (being the left-most points), and vice versa, point $B$ on circle $l$ corresponds to point $D$ on circle $r$ (being the right-most points). Let also point $I$ be the point of tangency of tangent line $t$ with circle $l$, and point $J$ be the point of tangency on circle $r$. It is the case then that lines $\overline{AI}$ and $\overline{CJ}$ are parallel, and indeed lines $\overline{BI}$ and $\overline{DJ}$ are also parallel. Now, proceed to the previous section to complete the construction.

This construction (and the previous one that is referenced) are used in the following section: constructing a center from two non-intersecting circles having only a centerline in common.

From Two Circles With A Common Centerline, or From Three Circles
This section is for the treatment of two non-intersecting circles devoid of their centers, but sharing a common centerline, or the scenario involving three non-intersecting circles. These constructions are rather complicated and requires a more thorough discussion. Many hundreds of line constructions are required, and the concepts are advanced, coming from projective geometry and including such topics as inversions, involutions, projections, harmonics, conjugates, homologies, fixed points, and others. Many preliminary constructions must be explained as they will be utilized many times, and they have yet to be discussed in either this subpage, or in the Poncelet-Steiner theorem main article. These are not trivial constructions.

In the two-circle scenario, one centerline common to both circles is already provided as a given to this scenario; to find a circle center we must only find a second centerline to either one of the two circles. In the three-circle scenario, many of the same concepts and constructions will carry over from the two-circle scenario. Refer to my second sandbox to continue the discussion.

Category:Euclidean plane geometry Category:Theorems in plane geometry Category:Compass and straightedge constructions