What is 4x^2+8x+4y^2-12y=51 ?

The solution to 4x^2+8x+4y^2-12y=51 is Circle with (a,b)=(-1, 3/2),r=4

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4x^2+8x+4y^2-12y=51

4 x^{2} + 8 x + 4 y^{2} - 12 y = 51

(a, b) = (- 1, \frac{3}{2}), r = 4

Solution steps

4 x^{2} + 8 x + 4 y^{2} - 12 y = 51

Rewrite

4 x^{2} + 8 x + 4 y^{2} - 12 y = 51

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{3}{2})^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{3}{2}), r = 4

x^{2} + y^{2} - 12 x = - 35

x^{2} + y^{2} + (- 6) + 4 y - 3 = 0

x^{2} - 14 x + y^{2} + 10 y + 58 = 0

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

4 x^{2} + 8 x + 4 y^{2} - 12 y = 23

What is 4x^2+8x+4y^2-12y=51 ?
The solution to 4x^2+8x+4y^2-12y=51 is Circle with (a,b)=(-1, 3/2),r=4