Angle of Intersection of Circles is Equal at Both Points

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 $\CC'$ at $A$ is equal to the angle of intersection of $\CC$ and $\CC'$ at $B$.

Proof

Consider the two triangles $CAC'$ and $CBC'$.

By definition of radius:

$CA = CB$ and $C'A = C'B$

\(\ds CA\)

\(=\)

\(\ds CB\)

Definition of Radius of Circle

\(\ds C'A\)

\(=\)

\(\ds C'B\)

Definition of Radius of Circle

\(\ds \leadsto \ \ \)

\(\ds \triangle CAC'\)

\(=\)

\(\ds \triangle CBC'\)

Triangle Side-Side-Side Equality

\(\ds \leadsto \ \ \)

\(\ds \angle CAC'\)

\(=\)

\(\ds \angle CBC'\)

The result follows from Angle of Intersection of Circles equals Angle between Radii.

$\blacksquare$

Sources

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