What is 4x^2+4y^2-4x+12y-6=0 ?

The solution to 4x^2+4y^2-4x+12y-6=0 is Circle with (a,b)=(1/2 ,-3/2),r=2

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4x^2+4y^2-4x+12y-6=0

4 x^{2} + 4 y^{2} - 4 x + 12 y - 6 = 0

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

Solution steps

4 x^{2} + 4 y^{2} - 4 x + 12 y - 6 = 0

Rewrite

4 x^{2} + 4 y^{2} - 4 x + 12 y - 6 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

(x^{2} - 14 x + 49) + (y^{2} + 8 y + 16) - 25 = 0

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

x^{2} + (y - 3)^{2} = 25

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

What is 4x^2+4y^2-4x+12y-6=0 ?
The solution to 4x^2+4y^2-4x+12y-6=0 is Circle with (a,b)=(1/2 ,-3/2),r=2