Q.Find the equation of the circle which passes through the origin and intersects the circles $x^2 + y^2 - 4x - 6y - 3 = 0$, $x^2 + y^2 - 8y + 12 = 0$ orthogonally.

Two circles are orthogonal (they cut each other at right angles) when the angle $\theta$ between them, as derived above, equals $90^\circ$, i.e. $\cos\theta=0$. Setting the numerator of the angle formula to zero,

gives the orthogonality condition for the circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g'x+2f'y+c'=0$. Geometrically, at the point of intersection of two orthogonal circles, each circle's tangent passes through the other circle's centre -- equivalently, the tangent from either centre to the other circle has length equal to that circle's own radius, which is exactly what the condition $d^2=r_1^2+r_2^2$ (the Pythagorean form of $\cos\theta=0$ in the cosine-rule identity) expresses. …