Question

Find the locus of the complex number $2$ satisfying $|z+2-3i|=7$

Answer 1

Given that a complex number ${}^{\prime }{z}^{\prime }$ satisfies

$|z+2-3i|=7⟶\left(1\right)$

let, $z=x+iy$, where $x,yϵR$

then, substituting in equation $\left(1\right)$ we get

$|x+iy+2-3i|=7$

$\Rightarrow |(x+2)+i(y-3)|=7$

$\Rightarrow \sqrt{{(x+2)}^2+{(y-3)}^2}=7$ $(\because |a+ib|=\sqrt{a^2+b^2})$

Squaring both the sides we get,

$\Rightarrow {(x+2)}^2+{(y-3)}^2=49$

$\Rightarrow x^2+4+4x+y^2+9-6y=49$

$\Rightarrow x^2+y^2+4x-6y-36=0$

$\therefore$ the locus of the complex number ${}^{\prime }{z}^{\prime }$ is a circle entered at $(-2,3)$ having a radius of $7$ units.