Question

$$x^{2}+y^{2}-16x-4y+32=0$$

Solve for x (complex solution)

$x=\sqrt{-\left(y-8\right)\left(y+4\right)}+8$
$x=-\sqrt{-\left(y-8\right)\left(y+4\right)}+8$

Solve for y (complex solution)

$y=\sqrt{\left(2-x\right)\left(x-14\right)}+2$
$y=-\sqrt{\left(2-x\right)\left(x-14\right)}+2$

Solve for x

$x=\sqrt{\left(8-y\right)\left(y+4\right)}+8$
$x=-\sqrt{\left(8-y\right)\left(y+4\right)}+8\text{, }y\geq -4\text{ and }y\leq 8$

Solve for y

$y=\sqrt{\left(14-x\right)\left(x-2\right)}+2$
$y=-\sqrt{\left(14-x\right)\left(x-2\right)}+2\text{, }x\geq 2\text{ and }x\leq 14$