Question

If $P,A,B$ represent the complex numbers $x+iy,6i$ and $3$ respectively and $P$ moves in such a manner that $PA=2PB$, then prove that $z\overline{z}=(4+2i)z+(4-2i)\overline{z}$. Also prove that the locus of the point $P$ is a circle whose centre is the point $(4-2i)$ and radius $\sqrt{20}$.

Answer 1

$P{A}^2=4P{B}^2$ $\Rightarrow$ ${|z-6i|}^2=4{|z-3|}^2$
Again from $\left(1\right)$, we have
$x^2+(y-6{)}^2=4[(x-3{)}^2+y^2]$
or $x^2+y^2-12y+36=4[x^2+y^2-6x+9]$
or $3(x^2+y^2)-24x+12y=0$
or $x^2+y^2-8x+4y=0$
Above represents a circle with center at $(4,-2)$ and radius $=v\sqrt{g^2+f^2-c}=\sqrt{20}=r$
In complex form, centre is $z_0=4-2i$ and $r=\sqrt{20}$
$\therefore$ ${|z-z_0|}^2=r^2$ or ${|z-(4-2i)|}^2=20$
Another form : $x^2+y^2-8x+4y=0$
$x^2+y^2=4\left(2x\right)+2i\left(2iy\right)$
${\left|z\right|}^2=4(z+\overline{z})+2i(z-\overline{z})$
or $z\overline{z}=(4+2i)z+(4-2i)\overline{z}$