Q.Find the equation of the circle which cuts the circles $x^2+y^2-4x-6y+5=0$, $x^2+y^2-2x-4y-1=0$ and $x^2+y^2+2x-2y-1=0$ orthogonally.

Step 1. From the previous problem, the radical centre of $S_1:x^2+y^2-4x-6y+5=0$, $S_2:x^2+y^2-2x-4y-1=0$, $S_3:x^2+y^2+2x-2y-1=0$ is $(-3,6)$.

Step 2. Compute the power of $(-3,6)$ with respect to $S_1$: $S_1(-3,6)=(-3)^2+6^2-4(-3)-6(6)+5=9+36+12-36+5=26$.

Step 3 (check with $S_2,S_3$). $S_2(-3,6)=9+36-2(-3)-4(6)-1=9+36+6-24-1=26$; $S_3(-3,6)=9+36+2(-3)-2(6)-1=9+36-6-12-1=26$. All three equal $26$, as required at the radical centre. …

Circle orthogonal to three circles: centred at their radical centre, with radius-squared equal to the common power of the radic …

• Using the distance from the radical centre to a circle's centre as the radius, instead of the tangent-length (power) value. …