Equation of Chord of Contact on Circle Centered at Origin

Theorem
Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.

Let $P = \tuple {x_0, y_0}$ be a point which is outside the boundary of $\CC$.

Let $UV$ be the chord of contact on $\CC$ with respect to $P$.

Then $UV$ can be defined by the equation:


 * $x x_0 + y y_0 = r^2$

Proof
Let $\TT_1$ and $\TT_2$ be a tangents to $\CC$ passing through $P$.

Let:
 * $\TT_1$ touch $\CC$ at $U = \tuple {x_1, y_1}$
 * $\TT_2$ touch $\CC$ at $V = \tuple {x_2, y_2}$

Then the chord of contact on $\CC$ with respect to $P$ is defined as $UV$.


 * Equation-of-Polar.png

From Equation of Tangent to Circle Centered at Origin, $\TT_1$ is expressed by the equation:


 * $x x_1 + y y_1 = r^2$

but as $\TT_1$ also passes through $\tuple {x_0, y_0}$ we also have:


 * $x_0 x_1 + y_0 y_1 = r^2$

This also expresses the condition that $U$ should lie on $\TT_1$:
 * $x x_0 + y y_0 = r^2$

Similarly, From Equation of Tangent to Circle Centered at Origin, $\TT_2$ is expressed by the equation:


 * $x x_2 + y y_2 = r^2$

but as $\TT_2$ also passes through $\tuple {x_0, y_0}$ we also have:


 * $x_0 x_2 + y_0 y_2 = r^2$

This also expresses the condition that $V$ should lie on $\TT_2$:
 * $x x_0 + y y_0 = r^2$

So both $U$ and $V$ lie on the straight line whose equation is:
 * $x x_0 + y y_0 = r^2$

and the result follows.