Wikipedia:Reference desk/Archives/Mathematics/2015 September 22

= September 22 =

Equation of line tangent to a circle
I just want to confirm my answer and my reasoning.

Find the linear equation for a line tangent to the circle x^2+y^2=169 at the point (-12,5).

The slope of that line will equal the x coordinate/y coordinate, but in this case it will be positive since the line is moving upwards. Then I just plug in that point into point slope form and I get 169/5 for the y-intercept. The final answer in point slope form would be y=12/5x+169/5 correct? ScienceApe (talk) 15:21, 22 September 2015 (UTC)


 * Let x=-12 into the equation and get y = -144/5+169/5 = 25/5 = 5. So the point (-12,5) is on the line. The vector (5,12) is along the line and perpendicular to (-12,5), so the line is a tangent to the circle. All right! Bo Jacoby (talk) 15:41, 22 September 2015 (UTC).


 * Since you have the slope of the line and a point on the line, you don't need to work out the y-intercept if you use the point-slope form of the line's equation: y &minus; 5 = (12/5)(x + 12). Add 5 to both side and you get the same answer you had above. --174.88.134.156 (talk) 01:36, 23 September 2015 (UTC)