Complex Number Question 10
Question 10 - 01 February - Shift 1
If the center and radius of the circle $|\frac{z-2}{z-3}|=2$ are respectively $(\alpha, \beta)$ and $\gamma$, then $3(\alpha+\beta+\gamma)$ is equal to
(1) 11
(2) 9
(3) 10
(4) 12
Show Answer
Answer: (4)
Solution:
Formula: Equation of a circle
$\sqrt{(x-2)^{2}+y^{2}}=2 \sqrt{(x-3)^{2}+y^{2}}$
$x^{2}+y^{2}-4 x+4=4 x^{2}+4 y^{2}-24 x+36$
$3 x^{2}+3 y^{2}-20 x+32=0$
$x^{2}+y^{2}-\frac{20}{3} x+\frac{32}{3}=0$
$(\alpha, \beta)=(\frac{10}{3}, 0)$
$\gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$
$3(\alpha+ \beta+ \gamma)=3(\frac{10}{3}+\frac{2}{3})=12$
