Circles - tangent from common point

Given the equation of a circle and the points of contact of two tangents, is it possible to find their point of intersection?

The obvious method is to find the equation of the two tangents, using the fact that they are perpendicular to the radius and pass through the point of contact, and then find their point of intersection. However, this method gets a bit lengthy. I was wondering if there is an easier method to do this? Or maybe an easier method to derive the equation of the tangents at least?

Outline: Draw a picture. Find the centre $C$ of the circle in the usual way.

If the given points of contact of the tangents are $A$ and $B$, find the coordinates of $M$, where $M$ is the midpoint of the line segment $AB$. For the midpoint, we average the $x$ coordinates, and average the $y$-coordinates.

Find the equation of the line $CM$. Straightforward, since we know two points on the line.

The intersection point of the two tangents is a point $X$ on this line such that the slope of $XA$ is the negative of the reciprocal of the known slope of $CA$.

I would say

The equation of the tangent to the circle $(x-m)^2+(y-n)^2=r^2$, in point $T(x_0,y_0)$ of circle has the form:

$(x-m)(x_0-m)+(y-n)(y_0-n)=r^2$

• I see..... and then the only way to find the point of intersection is to solve the two equations? Or is there a simple way for that too? – Gummy bears Aug 21 '14 at 15:59
• @Gummybears: I would say that the search for the intersection will always compute solutions of two equations. I do not think it can be appreciably simplified. – georg Aug 21 '14 at 16:45