What is 4x^2+4y^2-4y=7 ?

The solution to 4x^2+4y^2-4y=7 is Circle with (a,b)=(0, 1/2),r=sqrt(2)

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4x^2+4y^2-4y=7

4 x^{2} + 4 y^{2} - 4 y = 7

(a, b) = (0, \frac{1}{2}), r = 2

Solution steps

4 x^{2} + 4 y^{2} - 4 y = 7

Rewrite

4 x^{2} + 4 y^{2} - 4 y = 7

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, \frac{1}{2}), r = 2

x^{2} + y^{2} - 3 x - 4 y + 4 = 0

(x - \frac{5}{3})^{2} + (y + \frac{7}{3})^{2} = 64

x^{2} + y^{2} - 16 x + 4 y + 32 = 0

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

x^{2} - 4 x + y^{2} + 6 y = 12

What is 4x^2+4y^2-4y=7 ?
The solution to 4x^2+4y^2-4y=7 is Circle with (a,b)=(0, 1/2),r=sqrt(2)