Find the locus of this point P

Use vectors with $C$ at the origin. Then $$|\mathbf{PA}|^2+|\mathbf{PB}|^2=|\mathbf{PC}|^2\implies|\mathbf P|^2-2(\mathbf A+\mathbf B)\cdot\mathbf P+|\mathbf A|^2+|\mathbf B|^2=\mathbf 0.$$ Then if $\mathbf D=\mathbf A+\mathbf B$, this rearranges to $$|\mathbf P-\mathbf D|^2=2\mathbf A\cdot\mathbf B.$$ So as long as $\angle BCA<90^{\circ}$, the locus will be a circle with centre $D$ radius $\sqrt{2ab\cos C}$.

We extend median $CM$ such that $CM = MD$. Then please note that $MP$ is median of $\triangle CPD$ and $\triangle APB$. Applying Apollonius's theorem,

$PA^2 + PB^2 = 2(MP^2 + AM^2)$, $PC^2 + PD^2 = 2(MP^2 + CM^2)$

Subtracting second from first,

$PA^2 + PB^2 - PC^2 = 0 = PD^2 + 2 (AM^2 - CM^2)$

i.e $PD^2 + 2(\frac{c^2}{4} - \frac{1}{2}(a^2 + b^2) + \frac{c^2}{4}) = 0$

That leads to $PD = \sqrt{a^2+b^2-c^2} \ $. So the locus of point $P$ satisfying $PA^2 + PB^2 = PC^2$ is a circular arc with point $D$ being the center and radius $\sqrt{a^2+b^2-c^2}$. We also note that this is possible only if $a^2 + b^2 \geq c^2$.