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A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is
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Verified Answer
The correct answer is:
$x^2+y^2=36$
We have, length of median of $\triangle A B C=9$
$$
\therefore \quad A O=\frac{2}{3} A D \Rightarrow A O=\frac{2}{3} \times 9=6
$$
$O$ is the circumcentre of $\triangle A B C$.
We know that in equilateral triangle circumcentre, incentre, centroid coincide.
$\therefore \quad$ Origin $O(0,0)$ is the centre and $A O$ is radius of circle.
Hence, equation of circle
$$
\begin{aligned}
& (x-\sigma)^2+(y-\sigma)^2=(6)^2 \\
\Rightarrow & x^2+y^2=36
\end{aligned}
$$
$$
\therefore \quad A O=\frac{2}{3} A D \Rightarrow A O=\frac{2}{3} \times 9=6
$$
$O$ is the circumcentre of $\triangle A B C$.
We know that in equilateral triangle circumcentre, incentre, centroid coincide.
$\therefore \quad$ Origin $O(0,0)$ is the centre and $A O$ is radius of circle.
Hence, equation of circle
$$
\begin{aligned}
& (x-\sigma)^2+(y-\sigma)^2=(6)^2 \\
\Rightarrow & x^2+y^2=36
\end{aligned}
$$
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