What is 8x^2+8y^2-32x+64y-376=0 ?

The solution to 8x^2+8y^2-32x+64y-376=0 is Circle with (a,b)=(2,-4),r=sqrt(67)

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8x^2+8y^2-32x+64y-376=0

8 x^{2} + 8 y^{2} - 32 x + 64 y - 376 = 0

(a, b) = (2, - 4), r = 67

Solution steps

8 x^{2} + 8 y^{2} - 32 x + 64 y - 376 = 0

Rewrite

8 x^{2} + 8 y^{2} - 32 x + 64 y - 376 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 4))^{2} = (67)^{2}

Therefore the circle properties are:

(a, b) = (2, - 4), r = 67

4 x^{2} + 4 y^{2} - 4 x + 8 y + 1 = 0

(x + 3)^{2} + (y - \frac{4}{3})^{2} = 36

(x - 3)^{2} + (y - 5)^{2} = 4

(x - 1)^{2} + (y - 4)^{2} = 9

\frac{x ^{2}}{5} + \frac{y ^{2}}{5} = 53

What is 8x^2+8y^2-32x+64y-376=0 ?
The solution to 8x^2+8y^2-32x+64y-376=0 is Circle with (a,b)=(2,-4),r=sqrt(67)