Question

Find the equation in complex variables of all the circles which are orthogonal to $\left|z\right|=1$ and $|z-1|=4$.

Answer 1

The given circle are $x^2+y^2=1$ and $(x-1{)}^2+y^2=16$
or $x^2+y^2-1=0$ and $x^2+y^2-2x-15=0$.
Let the equation of the circle which cuts the above circles orthogonally be
$x^2+y^2+2gx+2fy+c=0$
Apply the condition $2g_1{g}_2+2f_1{f}_2=c-1+c_2$ with the given circles we get $c=1$ and $g=7$. Hence the family of circles is given by
$x^2+y^2+14x+2fy+1=0$
center $(-7,-f)$, $r=\surd (48+f^2)$
$(x+7{)}^2+(y+f{)}^2=(48+f^2)$
or $|z+(7+if)|=\surd (48+f^2)$
where $f$ is parameter.