What is 4x^2+4y^2-16x+4y-47=0 ?

The solution to 4x^2+4y^2-16x+4y-47=0 is Circle with (a,b)=(2,-1/2),r=4

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4x^2+4y^2-16x+4y-47=0

4 x^{2} + 4 y^{2} - 16 x + 4 y - 47 = 0

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

Solution steps

4 x^{2} + 4 y^{2} - 16 x + 4 y - 47 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 x + 4 y - 47 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} + y^{2} = 34

(x - 1)^{2} + (y + 2)^{2} = 16

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

x^{2} + y^{2} - 3 y = 12

What is 4x^2+4y^2-16x+4y-47=0 ?
The solution to 4x^2+4y^2-16x+4y-47=0 is Circle with (a,b)=(2,-1/2),r=4