What is 4x^2-8x+4y^2+28y=11 ?

The solution to 4x^2-8x+4y^2+28y=11 is Circle with (a,b)=(1,-7/2),r=4

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4x^2-8x+4y^2+28y=11

4 x^{2} - 8 x + 4 y^{2} + 28 y = 11

(a, b) = (1, - \frac{7}{2}), r = 4

Solution steps

4 x^{2} - 8 x + 4 y^{2} + 28 y = 11

Rewrite

4 x^{2} - 8 x + 4 y^{2} + 28 y = 11

in the form of the standard circle equation

(x - 1)^{2} + (y - (- \frac{7}{2}))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (1, - \frac{7}{2}), r = 4

(x + 3)^{2} + (y - 2)^{2} - 16 = 0

(X + 4)^{2} + (Y + 1)^{2} = 25

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

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

4 x^{2} + 12 x + 4 y^{2} + 16 - 11 = 0

What is 4x^2-8x+4y^2+28y=11 ?
The solution to 4x^2-8x+4y^2+28y=11 is Circle with (a,b)=(1,-7/2),r=4