Question

The area of an equilateral triangle inscribed in the circle $x^2+y^2+2gx+2fy+c=0$ is

• A. $\frac{3\sqrt{3}}{2}(g^2+f^2-c)$
• B. $\frac{3\sqrt{3}}{4}(g^2+f^2-c)$
• C. $\frac{3\sqrt{3}}{4}(g^2+f^2+c)$
• D. $\frac{3\sqrt{3}}{2}(g^2+f^2+c)$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{4}(g^2+f^2-c)$

Given circle is $x^2+y^2+2gx+2fy+c=0$
Let $C$ be its centre and $PQR$ be an equilateral triangle inscribed in the circle.

$C\equiv (-g,-f)$
Radius $CQ,$ $r=\sqrt{g^2+f^2-c}$
Since $\angle QPR={60}^{\circ }\Rightarrow \angle QCR={120}^{\circ }$
( $\because$ Angle subtended by an arc at the centre of circle is twice the angle subtended at its circumference.)
$\therefore \angle QCL=\frac{1}{2}\angle QCR=\frac{1}{2}\times {120}^{\circ }={60}^{\circ }$

From $△QLC,$
$QL=CQ\sin {60}^{\circ }=\frac{\sqrt{3}}{2}\sqrt{g^2+f^2-c}$
$\therefore QR=2\times QL=\sqrt{3}\sqrt{g^2+f^2-c}$

Now, area of $△PQR$
$=\frac{\sqrt{3}}{4}(QR{)}^2$
$=\frac{\sqrt{3}}{4}\times 3(g^2+f^2-c)$
$=\frac{3\sqrt{3}}{4}(g^2+f^2-c)$