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

Answer: (A)

Let $C=(x,y)$
Now, $C{A}^2=C{B}^2=A{B}^2$
$\Rightarrow {\left(x+a\right)}^2+y^2={\left(x-a\right)}^2+y^2={\left(2a\right)}^2$
$\Rightarrow x^2+2ax+a^2+y^2=4a^2...\left(i\right)$
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
$x^2+y^2-\frac{2a}{\sqrt{3}}y-a^2=0$
$\Rightarrow x^2+y^2-\frac{2\sqrt{3}ay}{3}-a^2=0$
$\Rightarrow 3^{2}2+3y^2-2\sqrt{3}ay=3a^2$