Q50 / 90 · JEE Main 2013 (22 Apr Online) · Mathematics
If two vertices of an equilateral triangle are $\mathrm{A}(-a, 0)$ and $B(a, 0), a>0$, and the third vertex C lies above $x$-axis then the equation of the circumcircle of $\triangle \mathrm{ABC}$ is :
- Correct answerA$3 x^2+3 y^2-2 \sqrt{3} a y=3 a^2$
- B$3 x^2+3 y^2-2 a y=3 a^2$
- C$x^2+y^2-2 a y=a^2$
- D$x^2+y^2-\sqrt{3} a y=a^2$
Solution
Let $\mathrm{C}=(x, y)$
Now, $\mathrm{CA}^2=\mathrm{CB}^2=\mathrm{AB}^2$
$$
\begin{aligned}
& \Rightarrow(x+a)^2+y^2=(x-a)^2+y^2=(2 a)^2 \\
& \Rightarrow x^2+2 a x+a^2+y^2=4 a^2 \\
& \text { and } x^2-2 a x+a^2+y^2=4 a^2
\end{aligned}
$$
From (i) and (ii), $x=0$ and $y=\pm \sqrt{3} a$
Since point $\mathrm{C}(x, y)$ lies above the $x$-axis and $a>0$, hence $y=\sqrt{3} a$
$$
\therefore \quad \mathrm{C}=(0, \sqrt{3} a)
$$
Let the equation of circumcircle be
$$
x^2+y^2+2 g x+2 f y+\mathrm{C}=0
$$
Since points A( $-a, 0), \mathrm{B}(a, 0)$ and $\mathrm{C}(0, \sqrt{3} a)$ lie on the circle, therefore
$$
\begin{aligned}
& a^2-2 g a+\mathrm{C}=0 \\
& a^2+2 g a+\mathrm{C}=0
\end{aligned}
$$
and $3 a^2+2 \sqrt{3} a f+\mathrm{C}=0$
From (iii), (iv), and (v)
$$
g=0, c=-a^2, f=-\frac{a}{\sqrt{3}}
$$
Hence equation of the circumcircle is
$$
\begin{aligned}
& x^2+y^2-\frac{2 a}{\sqrt{3}} y-a^2=0 \\
& \Rightarrow x^2+y^2-\frac{2 \sqrt{3} a y}{3}-a^2=0 \\
& \Rightarrow 3 x^2+3 y^2-2 \sqrt{3} a y=3 a^2
\end{aligned}
$$
Hence, the correct answer is option: 'A'.