Wikipedia:Reference desk/Archives/Mathematics/2017 August 25

= August 25 =

Revised Simultaneous quadratic equations
This follows on from my question yesterday.

In an xy plane, position four (distinct) points (x1,y1) (x2,y2) (x3,y3) (x0,y0).

Draw three circles centred at (x1,y1) (x2,y2) (x3,y3) passing through (x0,y0), with radii r1 r2 r3.

Given the 8 values x1 y1 x2 y2 x3 y3 (r2-r1) (r3-r1) (and you can have 0=(r1-r1) if you like as well), how can you calculate x0 and y0?

-- SGBailey (talk)
 * More. If it should happen that r1 = r2, then (r2-r1) will be 0 and you know that (x0,y0) will be on the perpendicular bisector of (x1,y1) (x2,y2). Generally (x3,y3) r3 will intersect with this line in two places. Does that mean to make a unique calculation I should add (x4,y4) r4 giving (r4-r1)?


 * Then imagine that (x1,y1) is (0,0) and (x2,y2) is (3,0). If (r2-r1) happens to equal 1, then one solution to this sub problem is (1,0) (r2-r1 = 2-1). What shape is the equivalent (to perpendicular bisector above) locus of satisfying points? Does this help?


 * -- SGBailey (talk)


 * This is basically the Problem of Apollonius. Dmcq (talk) 18:04, 25 August 2017 (UTC)