Tangents to Circle from Point are of Equal Length

Theorem

Let $\CC$ be a circle.

Let $P$ be a point in the exterior of $\CC$.

Let $PA$ and $PB$ be tangents to $\CC$ from $P$ touching $\CC$ at $A$ and $B$ respectively.

Then:

$PA = PB$

Proof

Let $O$ be the center of $\CC$.

Construct $OA$ and $OB$.

From Radius at Right Angle to Tangent:

$PA \perp OA$ and $PB \perp OB$

and so $\angle OAP = \angle OBP$ which equals a right angle.

Consider the right triangles $\triangle OAP$ and $\triangle OBP$. We have:

$OA = OB$ by definition of radius of circle

$\angle OAP = \angle OBP$

$OP$ is their common hypotenuse.

Hence by Triangle Right-Angle-Hypotenuse-Side Congruence:

$\triangle OAP \cong \triangle OBP$

and so $PA = PB$.

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): circle $(4) \ \text {(a)}$
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): circle $(4) \ \text {(a)}$