# Angle of Intersection of Circles equals Angle between Radii

## Theorem

Let $\CC$ and $\CC'$ be circles whose centers are at $C$ and $C'$ respectively.

Let $\CC$ and $\CC'$ intersect at $A$ and $B$.

The angle of intersection of $\CC$ and $\CC'$ is equal to the angle between the radii to the point of intersection.

## Proof

From Normal to Circle passes through Center, the straight line passing through the center of a circle is normal to that circle.

Hence the radii $CA$ and $C'A$ are perpendicular to the tangents to $\CC$ and $\CC'$ respectively.

Thus, with reference to the above diagram, we have that:

$\angle FAC = \angle DAC'$

as both are right angles.

Hence:

 $\ds \angle FAC'$ $=$ $\ds \angle FAC + \angle CAC'$ $\ds$ $=$ $\ds \angle FAD + \angle DAC'$ from above $\ds \leadsto \ \$ $\ds \angle FAC + \angle CAC'$ $=$ $\ds \angle FAD + \angle FAC$ as $\angle FAC = \angle DAC'$ $\ds \leadsto \ \$ $\ds \angle FAD$ $=$ $\ds \angle CAC'$

Hence the result.

$\blacksquare$