What is 4x^2+4y^2+4x-27=0 ?

The solution to 4x^2+4y^2+4x-27=0 is Circle with (a,b)=(-1/2 ,0),r=sqrt(7)

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4x^2+4y^2+4x-27=0

4 x^{2} + 4 y^{2} + 4 x - 27 = 0

(a, b) = (- \frac{1}{2}, 0), r = 7

Solution steps

4 x^{2} + 4 y^{2} + 4 x - 27 = 0

Rewrite

4 x^{2} + 4 y^{2} + 4 x - 27 = 0

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - 0)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, 0), r = 7

x^{2} + y^{2} + 6 x - 8 y + 12 = 0

4 (x^{2} - 2 x + 1) + 4 (y^{2} - y + 0) = 11 + 1 + 0

x^{2} + y^{2} + x + y = 85 - 7

x^{2} + y^{2} - 10 x + 12 y - 83 - 0 = 0

2 x^{2} + 2 y^{2} - 12 x + 8 y + 24 = 0

What is 4x^2+4y^2+4x-27=0 ?
The solution to 4x^2+4y^2+4x-27=0 is Circle with (a,b)=(-1/2 ,0),r=sqrt(7)