Q.Show that the circles $x^2+y^2-4x+2y+4=0$ and $x^2+y^2+6x-4y-20=0$ intersect 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.

A further consequence connects orthogonality to the radical axis (studied later in this chapter): if a circle cuts two other circles orthogonally, its centre must lie on the radical axis of those two circles. This is because the power of that centre with respect to each of the two circles equals the square of the corresponding tangent length, and orthogonality forces this tangent length to equal the cutting circle's own radius for both circles simultaneously -- so the two powers are equal, which is precisely the defining condition for a point to lie on the radical axis.