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