Question

If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x$ $-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}$ $+( y -2 \sqrt{2})^{2}= r ^{2}$, then the value of $r ^{2}$ is equal to

Answer 1

$PQ$ is diameter of circle
$S: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0 $
$C(\sqrt{2}, 3 \sqrt{2}), O(2 \sqrt{2}, 2 \sqrt{2}) $
$r_{1}=\sqrt{6}$
$S_{1}:(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} $
Now in$ \Delta OCQ$
$|O C|^{2}+|C Q|^{2}=|O Q|^{2} $
$4+6=r^{2}$
$r^{2}=10$