What is the area of the square with this circle inscribed in it?

Question

The circle $2x^2 = -2y^2 + 12x - 4y + 20$ is inscribed inside a square which has a pair of sides parallel to the x-axis. What is the area of the square?

I have manipulated it to get $$(x-3)^2+(y+1)^2=10.$$ The answer should be $(2\sqrt{10})^2,$ but this is incorrect. Help?

Answer 1

Because from your equation it should be $$\frac{(2\sqrt{10})^2}{2}=20.$$

Let $ABCD$ be a square.

Thus, $$S_{ABCD}=\frac{1}{2}AC\cdot BD.$$

The right equation of the circle it's $$(x-3)^2+(y+1)^2=20.$$ Can you end it now?

Answer 2

The original equation is $$2x^2 = -2y^2 + 12x - 4y + 20.$$ Moving terms around, we get $$2x^2 - 12x + 2y^2 + 4y = 20.$$ Dividing by $2$ gives $$x^2 - 6x + y^2 + 2y = 10.$$ Completing the square gives $$x^2 - 6x + 9 + y^2 + 2y + 1 = 10 + 9 + 1 = 20.$$ Factoring gives $$(x-3)^2 + (y+1)^2 = (\sqrt{20})^2.$$ I suspect your error could have been that you didn't add $10$ to both sides of the equation when you completed the square.