Length of Tangent from Point to Circle center Origin

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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:

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


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