What is 4x^2+4y^2-24y-13=0 ?

The solution to 4x^2+4y^2-24y-13=0 is Circle with (a,b)=(0,3),r= 7/2

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4x^2+4y^2-24y-13=0

4 x^{2} + 4 y^{2} - 24 y - 13 = 0

(a, b) = (0, 3), r = \frac{7}{2}

Solution steps

4 x^{2} + 4 y^{2} - 24 y - 13 = 0

Rewrite

4 x^{2} + 4 y^{2} - 24 y - 13 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 3)^{2} = (\frac{7}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 3), r = \frac{7}{2}

(x + 3)^{2} + (y + 2)^{2} = 16

x^{2} + 4 x + y^{2} - 4 y = 20

(x + 7)^{2} + (y + 5)^{2} = 50

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

x^{2} + y^{2} - 10 x - 18 y + 81 = 0

What is 4x^2+4y^2-24y-13=0 ?
The solution to 4x^2+4y^2-24y-13=0 is Circle with (a,b)=(0,3),r= 7/2