Circle Cutting Three Circles Orthogonally

The radical centre of three circles gives an immediate, direct construction of the unique circle that cuts all three of them orthogonally. Because the radical centre $R$ has equal power with respect to all three circles, computing that common power using any one of the three circles' equations (substituting $R$'s coordinates into it) gives a single value $t^2$ -- the square of the common tangent length from $R$ to each of the three circles.

Taking $R$ as centre and $t$ as radius produces the required circle. To see why this circle is orthogonal to (say) the first given circle, with centre $C_2$ and radius $r_2$: the orthogonality condition is $d^2=r_1^2+r_2^2$ where $d=RC_2$ and $r_1=t$. But $d^2-r_2^2$ is, by definition, exactly the power of $R$ with respect to that circle -- which equals $t^2=r_1^2$ by the very construction of $t$. So $d^2=r_1^2+r_2^2$ holds, confirming orthogonality; and since the same value $t^2$ was, by definition of the radical centre, also the power with respect to the other two given circles, the identical argument shows the constructed circle is orthogonal to all three simultaneously. …