A circle C₁ with equation x^2 + y^2 - 8x - 6y - 11 = 0 circumscribes an equilateral triangle. A second circle C₂ is inscribed in this equilateral triangle. If a square is inscribed in the circle C₂ , then the area of this square is

Step-by-step solution

The equation of circle C₁ is x^2 + y^2 - 8x - 6y - 11 = 0 . The radius of C₁ is R = g^2 + f^2 - c = (-4)^2 + (-3)^2 - (-11) = 16 + 9 + 11 = 6 . Since C₁ circumscribes the equilateral triangle, R = 6 is the circumradius of the triangle. For an equilateral triangle, the inradius r is half of the circumradius. Thus, the radius of the inscribed circle C₂ is r = R 2 = 3 . A square is inscribed in C₂ . The diagonal of this square is equal to the diameter of C₂ , which is 2r = 6 . The area of a square with diagonal d is d