Given two fixed points $A(-2,1)$ and $B(3,0)$, find the locus of a point $P$ which moves such that the angle $\angle APB$ is always a right angle.

Let, the coordinates of point $P\left(h, k\right)$

Given, points $A\left(-2, 1\right)$ and $B\left(3, 0\right)$,

So, the slope of $AP$ is (Say $m_1$) $\frac{k-1}{h+2}$,

and, the slope of $BP$ is (Say $m_2$ )$\frac{k-0}{h-3}=\frac{k}{h-3}$

As, $\angle APB=90°$

Hence, $m_1{m}_2=-1$

$\frac{k-1}{h+2}\times \frac{k}{h-3}=-1 \Rightarrow \frac{k^2-k}{h^2+2h-3h-6}=-1 \Rightarrow \frac{k^2-k}{h^2-h-6}=-1$

$k^2-k=-\left(h^2-h-6\right) \Rightarrow k^2+h^2-k-h-6=0$

So, replace $\left(h, k\right)\to \left(x, y\right)$, in order to obtain the locus.

Hence, the locus of point $P$ is $x^2+y^2-x-y-6=0$