A circle having centre at the origin passes through the three vertices of an equilateral triangle the length…

1. $x^2+y^2=9$
2. $x^2+y^2=18$
3. $x^2+y^2=36$
4. $x^2+y^2=81$

Solution

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} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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