What is 4x^2+4y^2+8x-16y-29=0 ?

The solution to 4x^2+4y^2+8x-16y-29=0 is Circle with (a,b)=(-1,2),r= 7/2

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4x^2+4y^2+8x-16y-29=0

4 x^{2} + 4 y^{2} + 8 x - 16 y - 29 = 0

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

Solution steps

4 x^{2} + 4 y^{2} + 8 x - 16 y - 29 = 0

Rewrite

4 x^{2} + 4 y^{2} + 8 x - 16 y - 29 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

(x - 2)^{2} + (y + 3)^{2} = 36

x^{2} + y^{2} + 16 y = - 39

x^{2} + y^{2} + 2 x - 6 y + 1 = 0

x^{2} + y^{2} = 8^{2}

What is 4x^2+4y^2+8x-16y-29=0 ?
The solution to 4x^2+4y^2+8x-16y-29=0 is Circle with (a,b)=(-1,2),r= 7/2