Question

If one of the diameters of the circle
$x^2+y^2−2\sqrt2x−6\sqrt2y+14=0$
is a chord of the circle
$(x−2\sqrt2)^2+(y−2\sqrt2)^2=r^2$
, then the value of r2 is equal to _______.

Answer 1

The correct answer is 10
For
$x^2+y^2−2\sqrt2x−6\sqrt2y+14=0$
Radius
$=\sqrt{(\sqrt2)^2+(3\sqrt2)^2−14}=\sqrt6$
⇒ Diameter $= 2\sqrt6$
If this diameter is chord to
$(x−2\sqrt2)^2+(y−2\sqrt2)^2=r^2$
then

Fig.

$⇒r^2=6+\left(\sqrt{(\sqrt2)^2+(\sqrt2)^2}\right)^2$
⇒ r2 = 6 + 4 = 10
⇒ r2 = 10