Question

If two vertices of an equilateral triangle are $A (- a, 0)$ and $B (a, 0), a > 0$, and the third vertex $C$ lies above x-axis then the equation of the circumcircle of $\Delta ABC$ is :

• A. $3x^{2}+3y^{2}-2\sqrt{3}ay=3a^{2}$
• B. $3x^{2}+3y^{2}-2ay=3a^{2}$
• C. $x^{2}+y^{2}-2ay=a^{2}$
• D. $x^{2}+y^{2}-\sqrt{3}ay=a^{2}$

Answer 1

The Correct Option is A

Let $C = (x, y)$
Now, $CA^2 = CB^2=AB^2$
$\Rightarrow \left(x + a\right)^{2}+y^{2} = \left(x- a\right)^{2}+y^{2} = \left(2a\right)^{2}$
$\Rightarrow x^{2} + 2ax + a^{2}+y^{2} = 4a^{2}\,...\left(i\right)$
and $x^{2} - 2ax + a^{2} +y^{2} = 4a^{2}\,...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right), x = 0$ and $y=\pm\sqrt{3}a$
Since point $C\left(x, y\right)$ lies above the x-axis and $a > 0$, hence $y=\sqrt{3}a$
$\therefore C=\left(0, \sqrt{3}a\right)$
Let the equation of circumcircle be
$x^2 +y^2 + 2gx + 2fy + C = 0$
Since points $A\left(- a, 0\right), B\left(a, 0\right)$ and $C\left(0, \sqrt{3}a\right)$ lie on the circle, therefore
$a^{2} - 2ga + C = 0\,...\left(iii\right)$
$a^{2} + 2ga + C = 0\,...\left(iv\right)$
and $3a^{2}+2\sqrt{3}af+C=0\,...\left(v\right)$
From $\left(iii\right), \left(iv\right)$, and $\left(v\right)$
$g=0, c=-a^{2}, f =-\frac{a}{\sqrt{3}}$
Hence equation of the circumcircle is
$x^{2}+y^{2}-\frac{2a}{\sqrt{3}}y-a^{2}=0$
$\Rightarrow x^{2}+y^{2}-\frac{2\sqrt{3}ay}{3}-a^{2}=0$
$\Rightarrow 3^{2}2 + 3y^{2}-2\sqrt{3}ay = 3 a^{2}$