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An equilateral triangle is inscribed in the circle $x^{2}+y^{2}=a^{2}$ with one of the vertices at $(a, 0)$. What is the equation of the $\begin{array}{ll}\text { side opposite to this vertex ? }
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Verified Answer
The correct answer is:
$2 \mathrm{x}+\mathrm{a}=0$
Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin.
So, $\frac{A O}{O D}=\frac{2}{1}$
and $\mathrm{OD}=\frac{1}{2} \mathrm{AO}=\frac{\mathrm{a}}{2}$
So, other vertices of triangle have coordinates,
$\left(-\frac{\mathrm{a}}{2}, \frac{\sqrt{3 \mathrm{a}}}{2}\right)$ and $\left[-\frac{\mathrm{a}}{2},-\frac{\sqrt{3}}{2} \mathrm{a}\right]$
$\left(-\frac{a}{2} \frac{\sqrt{3 a}}{2}\right) y$
$$
\left(-\frac{a}{2}, \frac{\sqrt{3 a}}{2}\right)
$$
$\therefore \quad$ Equation of line $\mathrm{BC}$ is
$$
\begin{aligned}
x &=-\frac{a}{2} \\
\Rightarrow \quad 2 x+a &=0
\end{aligned}
$$
So, $\frac{A O}{O D}=\frac{2}{1}$
and $\mathrm{OD}=\frac{1}{2} \mathrm{AO}=\frac{\mathrm{a}}{2}$
So, other vertices of triangle have coordinates,
$\left(-\frac{\mathrm{a}}{2}, \frac{\sqrt{3 \mathrm{a}}}{2}\right)$ and $\left[-\frac{\mathrm{a}}{2},-\frac{\sqrt{3}}{2} \mathrm{a}\right]$
$\left(-\frac{a}{2} \frac{\sqrt{3 a}}{2}\right) y$
$$
\left(-\frac{a}{2}, \frac{\sqrt{3 a}}{2}\right)
$$
$\therefore \quad$ Equation of line $\mathrm{BC}$ is
$$
\begin{aligned}
x &=-\frac{a}{2} \\
\Rightarrow \quad 2 x+a &=0
\end{aligned}
$$
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