Question

An equilateral triangle is inscribed in the circle $x^2+y^2-12x-8y+4=0$ then the length of the side is

• A. $2$
• B. $6$
• C. $4\sqrt{3}$
• D. $12$

Answer 1

The correct option is D $12$

$x^2+y^2-12x-8y+4=0$
Centre $=(6,4)$ and radius $=\sqrt{48}=4\sqrt{3}.$
Let $a$ be the side of an equilateral triangle
In $\Delta OAB$,
By cosine rule, for triangle

$\cos {120}^0=\frac{(4\sqrt{3}{)}^2+(4\sqrt{3}{)}^2-a^2}{2\times 4\sqrt{3}\times 4\sqrt{3}}$
$-1/2=\frac{2\times 48-a^2}{96}$
$-48=2\times 48-a^2$
$a^2=3\times 48$
$\Rightarrow a=12$