If the area of an equilateral triangle inscribed

Solution:

Let the sides of equilateral $\Delta$ inscribed in the circle be $a$, then

$\cos 30^{\circ}=\frac{a}{2 r}$

$\frac{\sqrt{3}}{2}=\frac{a}{2 r}$

$a=\sqrt{3} r$

Then, area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$

$=\frac{\sqrt{3}}{4}(\sqrt{3} r)^{2}$

$=\frac{3 \sqrt{3}}{4} r^{2}$

But it is given that area of equilateral triangle $=27 \sqrt{3}$

Then, $27 \sqrt{3}=\frac{3 \sqrt{3}}{4} r^{2}$

$r^{2}=36$

$\Rightarrow \quad r=6$

But $\left(-\frac{1}{2} \text { coeff. of } x\right)^{2}+\left(-\frac{1}{2} \text { coeff. of } y\right)^{2}$

$-$ constant term $=r^{2}$

$(-5)^{2}+(-6)^{2}-c=36$