What is (4x^2-56)+4y^2+8y=56 ?

The solution to (4x^2-56)+4y^2+8y=56 is Circle with (a,b)=(0,-1),r=sqrt(29)

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(4x^2-56)+4y^2+8y=56

(4 x^{2} - 56) + 4 y^{2} + 8 y = 56

(a, b) = (0, - 1), r = 29

Solution steps

4 x^{2} - 56 + 4 y^{2} + 8 y = 56

Rewrite

4 x^{2} - 56 + 4 y^{2} + 8 y = 56

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = (29)^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = 29

X^{2} + Y^{2} + 2 X - 3 Y = 0

x^{2} + (y - \frac{7}{2})^{2} = 81

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

x^{2} + y^{2} - 14 x - 8 y = 0

x^{2} + y^{2} = 4 y + 3

What is (4x^2-56)+4y^2+8y=56 ?
The solution to (4x^2-56)+4y^2+8y=56 is Circle with (a,b)=(0,-1),r=sqrt(29)