Length of Tangent from Point to Circle center Origin
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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 |
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $11$. Geometrical meaning of the expression $x^2 + y^2 - r^2$