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

Correct Answer: 10

To solve the problem, we begin by analyzing the equation of the first circle: \(x^2+y^2−2\sqrt{2}x−6\sqrt{2}y+14=0\).
It's in the expanded form of a circle equation: \(x^2+y^2+Dx+Ey+F=0\). We complete the square for \(x\) and \(y\).
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