Question

If the center and radius of the circle $\left|\frac{z-2}{z-3}\right|=2$ are respectively $(\alpha, \beta)$ and $\gamma$, then $3(\alpha+\beta+\gamma)$ is equal to

• A. 12
• B. 11
• C. 10
• D. 9

Answer 1

Answer: (A)

$ \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} $
$ =x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 $
$ =3 x^2+3 y^2-20 x+32=0 $
$ =x^2+y^2-\frac{20}{3} x+\frac{32}{3}=0 $
$ =(\alpha, \beta)=\left(\frac{10}{3}, 0\right) $
$\gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} $
$3(\alpha, \beta, \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right)$
$=12$