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

Answer: 10

$PQ$ 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$\Delta OCQ$
$|OC{|}^2+|CQ{|}^2=|OQ{|}^2$
$4+6=r^2$
$r^2=10$