# Equation of Normal to 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 $P = \tuple {x_1, y_1}$ be a point on $\CC$.

Let $\NN$ be a normal to $\CC$ passing through $P$.

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

$y_1 x - x_1 y = 0$

## Proof

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

From Equation of Tangent to Circle Centered at Origin, $\TT$ can be described using the equation:

$x x_1 + y y_1 = r^2$

expressible as:

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

where the slope of $\TT$ is $-\dfrac {x_1} {y_1}$.

By definition, the normal is perpendicular to the tangent.

From Condition for Straight Lines in Plane to be Perpendicular, the slope of $\NN$ is $\dfrac {y_1} {x_1}$

Hence the equation for $\NN$ is:

 $\ds y - y_1$ $=$ $\ds \dfrac {y_1} {x_1} \paren {x - x_1}$ $\ds \leadsto \ \$ $\ds x_1 \paren {y - y_1}$ $=$ $\ds y_1 \paren {x - x_1}$ $\ds \leadsto \ \$ $\ds x_1 y$ $=$ $\ds y_1 x$

Hence the result.

$\blacksquare$