Question

Let $z_1 = 2 + 3i$ and $z_2 = 3 + 4i$ be two points on the complex plane. Then the set of complex numbers $z$ satisfying $|z - z_1|^2 + |z - z_2|^2 = |z_1 - z^2|^2$ represents

• A. a straight line
• B. a point
• C. a circle
• D. a pair of straight lines

Answer 1

Answer: (C)

Given, $z_{1}=2+3 i$ and $z_{2}=3+4 i$
Now, we have
$\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2} $ (let $\left.z=x+i y\right)$
$\Rightarrow |(x+i y)-(2+3 i)|^{2}+|(x+i y)-(3+4 i)|^{2}$
$=|(2+3 i)-(3+4 i)|^{2}$
$\Rightarrow |(x-2)+i(y-3)|^{2}+|(x-3)+i(y-4)|^{2}$
$=|-1-i|^{2}$
$\Rightarrow (x-2)^{2}+(y-3)^{2}+(x-3)^{2}+(y-4)^{2}=1+1$
$\Rightarrow x^{2}+4-4 \,x+y^{2}+9-6\, y$
$+x^{2}+9-6 x+y^{2}+16-8 y=2$
$\Rightarrow 2 x^{2}+2 \,y^{2}-10\, x-14 \,y+36=0$
$\Rightarrow x^{2}+y^{2}-5 \,x-7 \,y+18=0$
which represent a circle with centre $\left(\frac{5}{2,} \frac{7}{2}\right)$ and
radius $\sqrt{\frac{25}{4}+\frac{49}{4}-18}=\frac{1}{\sqrt{2}}$