find the area of the equilateral triangle inscribed in a circle $x^2+y^2+2gx+2fy+c=0$ ?

Question

I am stuck on the following problem:

How can I find the area of the equilateral triangle inscribed in a circle $x^2+y^2+2gx+2fy+c=0$ ?

Answer is given to be : $\frac{3\sqrt 3}{4}(g^2+f^2-c)$ sq. unit.

Can someone please explain?

Answer 1

Hint: First complete the square, giving an equation like $(x-h)^2+(y-k)^2 = r^2$. From this you can deduce the radius of the circle; then apply known formulas to figure out the area of the inscribed equilateral triangle (if you don't know these, look here for example).

Answer 2

In brief: Complete the square in your equation.

$$x^2+y^2+2gx+2fy+c=0$$ $$x^2+2gx+y^2+2fy=-c$$ $$x^2+2gx+g^2+y^2+2fy+f^2=g^2+f^2-c$$ $$(x+g)^2+(y+f)^2=\left(\sqrt{g^2+f^2-c}\right)^2$$

This is the standard form of an equation of a circle. We see that the radius of the circle is $\sqrt{g^2+f^2-c}$.

Now use that to find the area of the inscribed circle. That is straightforward geometry: let us know if you have difficulty with that.