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:

\(\ds {{{l|}}}\)

\(=\)

\(\ds \dfrac {y_1} {x_1} \paren {x - x_1}\)

\(\ds \leadsto \ \ \)

\(\ds {{{l|}}}\)

\(=\)

\(\ds y_1 \paren {x - x_1}\)

\(\ds \leadsto \ \ \)

\(\ds {{{l|}}}\)

\(=\)

\(\ds y_1 x\)

Hence the result.

$\blacksquare$

Sources

• 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $3$. The normal