Question

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

Answer 1

REF.Image

Center (p.q) radius = r

Then equation of circle is

$(x-p{)}^2+(y-q{)}^2=r^2$

Given equation

$x^2+y^2-6x+4y+5=0$

$x^2-6x+9+y^2+4y+4-8=0$

$(x-3{)}^2+(y+2{)}^2=8$

$p=3$ q=-2 $r^2=8$

$\therefore r=2\sqrt{2}$

equilateral triangle of side a

$sin60=\frac{BM}{OB}$ $\because BM=OBsin60$

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

$=\sqrt{6}$

$\therefore BC=28M$

$=2\sqrt{6}$ = side of equilateral $\Delta$

$\therefore$Area of triangle = $\frac{\sqrt{3}}{4}\times (side{)}^2$

$=\frac{\sqrt{3}}{4}\times 4\times 6$

$=6\sqrt{3}$ sq units.