The circumcentre of the triangle with vertices at $A\left(5,12\right),B\left(12,5\right),C\left(2\sqrt{13},3\sqrt{13}\right)$ is

Given vertices $A\left(5,12\right),B\left(12,5\right),C\left(2\sqrt{13},3\sqrt{13}\right)$  let $S=\left(x_1\right)$

$SA=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\sqrt{{\left(x-5\right)}^2+{\left(y-12\right)}^2}\to \left(1\right)$

$SB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\sqrt{{\left(x-12\right)}^2+{\left(y-5\right)}^2}\to \left(2\right)$

Now equate $\left(1\right)=\left(2\right)$

$\sqrt{{\left(x-5\right)}^2+{\left(y-12\right)}^2}=\sqrt{{\left(x-12\right)}^2+{\left(y-5\right)}^2}$

Now squaring on both sides

${\left(x-5\right)}^2+{\left(y-12\right)}^2={\left(x-12\right)}^2+{\left(y-5\right)}^2$

$x^2+25-10x+y^2+144-24y=x^2+144-24x+y^2+25-10y$

$14x-14y=0$

$14x=14y$

$\boxed{x=y}$

$SA=\sqrt{{\left(x-5\right)}^2+{\left(y-12\right)}^2}\to \left(3\right)$

We know that,

Now $SC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\sqrt{{\left(x-2\sqrt{13}\right)}^2+{\left(y-3\sqrt{13}\right)}^2}\to \left(4\right)$

Now equate $\left(3\right)=\left(4\right)$

$\sqrt{{\left(x-5\right)}^2+{\left(y-12\right)}^2}=\sqrt{{\left(x-2\sqrt{13}\right)}^2+{\left(y-3\sqrt{13}\right)}^2}$

Now squaring on both sides

${\left(x-5\right)}^2+{\left(y-12\right)}^2={\left(x-2\sqrt{13}\right)}^2+\left(y-3\sqrt{13}\right)$

$x^2+25-10x+y^2+144-24y=x^2+52-4\sqrt{13}x+y^2+117-6\sqrt{13}y$

$169-10x-24y=169-4\sqrt{13}x-6\sqrt{13}y$

$4\sqrt{13}x-10x=24y-6\sqrt{13}y$

Now put (1) and (2) solved equation

$\boxed{x=4}$

$4\sqrt{13}y-10y=24y-6\sqrt{13}y$

$y\left(4\sqrt{13}-10\right)=y\left(24-6\sqrt{13}y\right)$

$y\left(4\sqrt{13}-10-24+6\sqrt{13}\right)=0$

$\boxed{y=0}\boxed{\left(x,y\right)=\left(0,0\right)}$

Now $y=0$  means $x=0$

Then the circumcentre $S\left(x,y\right)=\left(0,0\right)$