Angle of Intersection of Two Circles

Angle θ between two circles at a point of intersection, defined as the angle between their tangents (equivalently, their radii) at that point.

The angle of intersection of two circles at a common point $P$ is defined as the angle between their tangents at $P$, which equals the angle between their radii at $P$.

Let the circles have centres $O_1$, $O_2$, radii $r_1$, $r_2$, and distance $d = O_1O_2$.

In triangle $O_1PO_2$, the sides are $r_1$, $r_2$, and $d$. The angle $\theta$ at $P$ (between the two radii, hence the angle of intersection) satisfies the cosine rule:

$$d^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\theta$$ $$\cos\theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$$

Equivalently:

$$\cos\theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2} \quad \text{(with sign reversed, depending on the angle taken)}$$

In terms of general form coefficients: For $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$:

$$d^2 = (g_1-g_2)^2 + (f_1-f_2)^2, \quad r_1^2 = g_1^2+f_1^2-c_1, \quad r_2^2 = g_2^2+f_2^2-c_2$$

Substituting:

$$\cos\theta = \frac{2g_1g_2 + 2f_1f_2 - c_1 - c_2}{2\sqrt{(g_1^2+f_1^2-c_1)(g_2^2+f_2^2-c_2)}}$$

The special case $\theta = 90°$ gives $\cos\theta = 0$, which is the orthogonality condition.