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 ____________.

For x^2+y^2-2 2 x-6 2 y+14=0 Radius =(2)^2+(3 2)^2-14=6 Diameter =2 6 If this diameter is chord to aligned &(x-2 2)^2+(y-2 2)^2=r^2 then \\ &&n

If one of the diameters of the circle x2 y2 2 2 x 6 2 y 14 0: (Mathematics)

For $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$

$\text { Radius }=\sqrt{(\sqrt{2})^{2}+(3 \sqrt{2})^{2}-14}=\sqrt{6}$

$\Rightarrow$ Diameter $=2 \sqrt{6}$

If this diameter is chord to

$\begin{aligned} &(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\ &\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\ &\Rightarrow r^{2}=6+4=10 \\ &\Rightarrow r^{2}=10 \end{aligned}$

Explanation

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