Subtract (1) from (2) to get a linear equation in $x,y$
Subtract (1) from (3) to get another
From these solve for $x$ and $y$.
Substitute back in (1) to get a quadratic in $h^2$.
Solve to get $h^2$
Note till the very end, you can call $z=h^2$ and finally solve for $h$
Added in response to comments
Here is a worked out example (the following is computer generated so excuse the poor formatting).
Let the equations be
$$\left(y-6\right)^2+\left(x-9\right)^2-{{5\,h^2}\over{64}} \tag 1
$$
$$
\left(y-9\right)^2+\left(x-19\right)^2-{{61\,h^2}\over{32}} \tag2
$$
$$ \left(
y-7\right)^2+\left(x-1\right)^2-{{25\,h^2}\over{32}} \tag 3$$
Subtract (1) from (2) and also (1) from (3) and simplify to get
$$ -384\,y-1280\,x-117\,h^2+20800 =0 \\ -128\,y+1024\,x-45\,h^2-
4288 =0 $$
Solving the two equations, letting $z=h^2$ we get
$$x={{9\,z+16832}\over{2176}} , ~~~y=-{{693\,z-61760}\over{2176}} \tag 4
$$
Now substitute in (1) to get
$$240165\,z^2-33961600\,z+1189826560=0$$
Solving
$$ z={{3718208}\over{48033}} , z=64 $$
which gives
$$
h = \pm {{8\,\sqrt{58097}}\over{9\,\sqrt{593}}}, h=\pm 8$$
Substituting in (4) we get
$$x={{42992}\over{5337}} , y={{19904}\over{5337}} , h=-
{{8\,\sqrt{58097}}\over{9\,\sqrt{593}}} \\ x={{42992
}\over{5337}} , y={{19904}\over{5337}} h={{8\,\sqrt{58097}}\over{9
\,\sqrt{593}}} \\x=8 , y=8 , h=8 \\ x=
8 , y=8 , h=-8 $$