- Home
- Standard 11
- Mathematics
10-1.Circle and System of Circles
hard
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
A
$15$
B
$70$
C
$18$
D
$10$
(JEE MAIN-2022)
Solution
$S: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$
$C(\sqrt{2}, 3 \sqrt{2}), O(2 \sqrt{2}, 2 \sqrt{2})$
$r_{1}=\sqrt{6}$
$S_{1}:(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}$
$\text { Now in } \Delta OCQ$
$|O C|^{2}+|C Q|^{2}=|O Q|^{2}$
$4+6=r^{2}$
$r^{2}=10$
Standard 11
Mathematics
