# Length of Tangent from Point to Circle center Origin

## Theorem

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

Let $P = \tuple {x, y}$ be a point in the plane of $\CC$ in the exterior of $\CC$.

Let $PT$ be a tangent to $\CC$ from $P$ such that $T$ is the point of tangency.

Then the length of $PT$ is given by:

$PT^2 = x^2 + y^2 - r^2$

## Proof

Let $\NN$ be the normal to $\CC$ at the point $T$.

From Normal to Circle passes through Center, $\NN$ passes through $O$.

By definition of the normal to $\CC$, $\NN$ is perpendicular to $PT$.

Hence $OT$, $PT$ and $OP$ form a right triangle whose hypotenuse is $OP$.

As $OT$ is a line segment coinciding with a radius of $\CC$:

$OT = r$

The length of $OP$ comes from the Distance Formula:

$OP = \sqrt {\paren {x - 0}^2 + \paren {y - 0}^2} = \sqrt {x^2 + y^2}$

Hence we have:

 $\ds OP^2$ $=$ $\ds OT^2 + PT^2$ Pythagoras's Theorem $\ds \leadsto \ \$ $\ds PT^2$ $=$ $\ds x^2 + y^2 - r^2$ substituting for $OP$ and $OT$ from above

$\blacksquare$