Question

If $\frac{|z-2|}{|z-3|}=2$ represents a circle, then its radius is equal to

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

Given, $\frac{|z-2|}{|z-3|}=2$
$\Rightarrow$ $|z-2|=2|z-3|$
$\Rightarrow$ $\sqrt{{(x-2)}^2+y^2}=2\sqrt{{(x-3)}^2+y^2}$
$\Rightarrow$ ${(x-2)}^2+y^2=4[{(x-3)}^2+y^2]$ (on squaring both sides)
$\Rightarrow$ $x^2+y^2-4x+4=4x^2+4y^2+36-24x$
$\Rightarrow$ $3x^2+3y^2-20x+32=0$
or $x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0$ ... (i)
We know that, standard equation of a circle is
$x^2+y^2+2gx+2fy+c=0$ ... (ii)
On comparing Eqs. (i) and (ii), we get
$2g=\frac{-20}{3}\Rightarrow g=\frac{-10}{3},f=0,c=\frac{32}{3}$
Hence, radius $=\sqrt{g^2+f^2-c}$
$=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$