JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 22)
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 r2 is equal to ____________.
Answer
10
Explanation
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
$$ \begin{aligned} &(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\\\ &\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\\\ &\Rightarrow r^{2}=6+4=10 \\\\ &\Rightarrow r^{2}=10 \end{aligned} $$
$$ \text { Radius }=\sqrt{(\sqrt{2})^{2}+(3 \sqrt{2})^{2}-14}=\sqrt{6} $$
$\Rightarrow$ Diameter $=2 \sqrt{6}$
If this diameter is chord to
$$ \begin{aligned} &(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\\\ &\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\\\ &\Rightarrow r^{2}=6+4=10 \\\\ &\Rightarrow r^{2}=10 \end{aligned} $$
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