Question

If $\left|\frac{z-2}{z-3}\right|=2$ represents a circle whose radius is $r$, then the value of $3r$ is

Answer 1

Let $z=x+iy$
$\left|\frac{z-2}{z-3}\right|=2\Rightarrow |z-2{|}^2=4|z-3{|}^2\Rightarrow (x-2{)}^2+y^2=4[(x-3{)}^2+y^2]\Rightarrow 3x^2+3y^2-24x+4x+36-4=0\Rightarrow x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0$
Radius of the circle is
$r=\sqrt{\frac{100}{9}-\frac{32}{3}}\Rightarrow r=\frac{2}{3}\therefore 3r=2$