The area of an equilateral triangle inscribed in the circle $x^2+y^2-6 x+2 y-28=0$ is sq. units

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Aicharya · Accepted Answer

Equation of given circle is $\begin{aligned} & & x^2+y^2-6 x+2 y-28 & =0 \ \Rightarrow & & (x-3)^2+(y+1)^2 & =38 \end{aligned}$ Now, area of equilateral triangle $\begin{aligned} A B C & =3~(\text {Area of } \triangle G B C) \ & =3\left(\frac{1}{2} r^2 \sin 120^{\circ}\right)=\frac{3}{2}(38) \frac{\sqrt{3}}{2}=\frac{57 \sqrt{3}}{2} \end{aligned}$

The area of an equilateral triangle inscribed in the circle \(x^2+y^2-6 x+2 y-28=0\) is sq. units

Solution