Angle of Intersection of Circles equals Angle between Radii

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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.


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.


\(\ds \angle FAC'\) \(=\) \(\ds \angle FAC + \angle CAC'\)
\(\ds \) \(=\) \(\ds \angle FAD + \angle DAC'\) 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.