# Angle of Intersection of Circles equals Angle between Radii

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## 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.

$\blacksquare$

## Sources

- 1933: D.M.Y. Sommerville:
*Analytical Conics*(3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $15$. Angle of intersection of two circles