Question

If $z_1=10+6i,z_2=4+2i$ such that $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{4},$ then find the centre and radius of the locus of complex number z.

• A. center is $(7,5)$ and radius is $\sqrt{26}$
• B. center is $(5,7)$ and radius is $\sqrt{26}$
• C. center is $(5,7)$ and radius is $\sqrt{13}$
• D. center is $(7,5)$ and radius is $\sqrt{13}$

Answer 1

The correct option is B center is $(5,7)$ and radius is $\sqrt{26}$

$r=\sqrt{26},C(5+7i).$
$\overrightarrow{AB}=6+4i\therefore \left|AB\right|=\sqrt{36+16}=\sqrt{52}$
If $C\left(z_0\right)$ be the centre, then angle at $C=2\left(\frac{\pi }{4}\right)=\frac{\pi }{2}.$
If r be the radius then $r^2+r^2=\sqrt{52}$
$\therefore r=\sqrt{26}.$
Again by rotation through an angle of $\pi /2$ in anti-clockwise direction,
$z_1-z_0=(z_2-z_0)e^{i\pi /2}=i(z_2-z_0)$
$\therefore z_1-i{z}_2=z_0(1-i)$
$z_0=\frac{z_1-i{z}_2}{1-i}=\frac{(1+i)(z_1-i{z}_2)}{2}$
$=\frac{1+i}{2}[(10+6i)-i(4+2i)]$
$=(1+i)[6+i]=(6-1)+i(6+1)=5+7i$
Ans: B