# Equation of Tangent to Circle Centered at Origin/Proof 2

## Theorem

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

Let $P = \tuple {x_1, y_1}$ be a point on $\CC$.

Let $\TT$ be a tangent to $\CC$ passing through $P$.

Then $\TT$ can be defined by the equation:

$x x_1 + y y_1 = r^2$

## Proof

From the slope-intercept form of a line, the equation of a line passing through $P$ is:

$y - y_1 = \mu \paren {x - x_1}$

If this line passes through another point $\tuple {x_2, y_2}$ on $\CC$, the slope of the line is given by:

$\mu = \dfrac {y_2 - y_1} {x_2 - x_1}$

Because $P$ and $Q$ both lie on $\CC$, we have:

 $\ds x_1^2 + y_1^2$ $=$ $\ds r^2 = x_2^2 + y_2^2$ $\ds \leadsto \ \$ $\ds y_2^2 - y_1^2$ $=$ $\ds x_2^2 - x_1^2$ $\ds \leadsto \ \$ $\ds \dfrac {y_2 - y_1} {x_2 - x_1}$ $=$ $\ds -\dfrac {x_1 + x_2} {y_1 + y_2}$

As $Q$ approaches $P$, we have that $y_2 \to y_1$ and $x_2 \to x_1$.

The limit of the slope is therefore:

$-\dfrac {2 x_1} {2 y_1} = -\dfrac {x_1} {y_1}$

The equation of the tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$ is therefore:

 $\ds y - y_1$ $=$ $\ds -\dfrac {x_1} {y_1} \paren {x - x_1}$ $\ds \leadsto \ \$ $\ds y_1 \paren {y - y_1}$ $=$ $\ds -x_1 \paren {x - x_1}$ $\ds \leadsto \ \$ $\ds x x_1 + y y_1$ $=$ $\ds x_1^2 + y_1^2$ $\ds$ $=$ $\ds r^2$ as $\tuple {x_1, y_1}$ is on $\CC$

$\blacksquare$