If one of the diameters of the circle x 2 + y 2 - 2 2 x - 6 2 y + 14 = 0 is a chord of the circle x - 2 2 2 + y - 2 2 2 = r 2 , then the value of r 2 is equal to

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Given that, one of the diameter of the circle ${(x - \sqrt{2})}^{2} + {(y - 3 \sqrt{2})}^{2} = 6$ is a chord of another circle ${(x - 2 \sqrt{2})}^{2} + {(y - 2 \sqrt{2})}^{2} = r^{2}$ .

Let, $C$ and $C_{1}$ be the centres of the circles ${(x - \sqrt{2})}^{2} + {(y - 3 \sqrt{2})}^{2} = 6$ and ${(x - 2 \sqrt{2})}^{2} + {(y - 2 \sqrt{2})}^{2} = r^{2}$ respectively.

So, $C (\sqrt{2}, 3 \sqrt{2}) & C_{1} (2 \sqrt{2}, 2 \sqrt{2})$

The position of the two circles as shown in figure.

Now, $C C_{1} = \sqrt{{(2 \sqrt{2} - \sqrt{2})}^{2} + {(2 \sqrt{2} - 3 \sqrt{2})}^{2}} = \sqrt{2 + 2} = 2$

It is clear from the above diagram, $∆ A C C_{1}$ is a right angled triangle. $\Rightarrow r^{2} = {(A C)}^{2} + {(C C_{1})}^{2} = {(\sqrt{6})}^{2} + 2^{2}$ $\Rightarrow r^{2} = 10$

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