What is 4x^2+4y^2-4x+12y-71=0 ?

The solution to 4x^2+4y^2-4x+12y-71=0 is Circle with (a,b)=(1/2 ,-3/2),r= 9/2

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4x^2+4y^2-4x+12y-71=0

4 x^{2} + 4 y^{2} - 4 x + 12 y - 71 = 0

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = \frac{9}{2}

Solution steps

4 x^{2} + 4 y^{2} - 4 x + 12 y - 71 = 0

Rewrite

4 x^{2} + 4 y^{2} - 4 x + 12 y - 71 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{9}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = \frac{9}{2}

x, x^{2} + Y (x)^{2} = 64

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

\frac{x ^{2}}{2} + \frac{y ^{2}}{2} = 2

x^{2} - 4 x + y^{2} - 9 x + 9 = 0

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

What is 4x^2+4y^2-4x+12y-71=0 ?
The solution to 4x^2+4y^2-4x+12y-71=0 is Circle with (a,b)=(1/2 ,-3/2),r= 9/2