Perpendicular bisector of $A\, (1,3)$ and $B\,(-3,5)$ is
$2 x\left(x_{1}-x_{2}\right)+2 y\left(y_{1}-y_{2}\right)=\left(x_{1}^{2}+y_{1}^{2}\right)-\left(x_{2}^{2}+y_{2}^{2}\right)$
$\Rightarrow 2 x(1+3)+2 y(3-5)=(1+9)-(9+25)$
$\Rightarrow 8 x-4 y=10-34$
$\Rightarrow 2 x-y+6=0 \,\,\,\,\ldots(i)$
Similarly, perpendicular bisector of $A(1,3)$ and $C(5,-1)$ is
$2 x(1-5)+2 y(3+1)=(1+9)-(25+1)$
$\Rightarrow -8 x+8 y=10-26$
$\Rightarrow -x+y=-2$
$\Rightarrow x-y-2=0 \,\,\,\,\ldots(ii)$
Point of intersection of Eqs. $(i)$ and $(ii)$, is
$P=(-8,-10)$.
Then, the distance between $P$ and $A$ is
$P A=\sqrt{(1+8)^{2}+(3+10)^{2}}$
$=\sqrt{81+169}$
$=\sqrt{250}=5 \sqrt{10}$
