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 (integer)
10
Solution
For $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$
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$\text { Radius }=\sqrt{(\sqrt{2})^{2}+(3 \sqrt{2})^{2}-14}=\sqrt{6}$
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$\Rightarrow$ Diameter $=2 \sqrt{6}$
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If this diameter is chord to
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$$
\begin{aligned}
&(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\\\
&\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\\\
&\Rightarrow r^{2}=6+4=10 \\\\
&\Rightarrow r^{2}=10
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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