What is 16x^2+16y^2+96-40y-315=0 ?

The solution to 16x^2+16y^2+96-40y-315=0 is Circle with (a,b)=(0, 5/4),r=(sqrt(61))/2

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16x^2+16y^2+96-40y-315=0

16 x^{2} + 16 y^{2} + 96 - 40 y - 315 = 0

(a, b) = (0, \frac{5}{4}), r = \frac{61}{2}

Solution steps

16 x^{2} + 16 y^{2} + 96 - 40 y - 315 = 0

Rewrite

16 x^{2} + 16 y^{2} + 96 - 40 y - 315 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - \frac{5}{4})^{2} = (\frac{61}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, \frac{5}{4}), r = \frac{61}{2}

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

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

(x + 4)^{2} + (y + 12)^{2} - 16 = 0

x^{2} + y^{2} + 6 x + 16 y - 8 = 0

4 x^{2} + 4 y^{2} + 8 x - 4 y - 3 = 0

What is 16x^2+16y^2+96-40y-315=0 ?
The solution to 16x^2+16y^2+96-40y-315=0 is Circle with (a,b)=(0, 5/4),r=(sqrt(61))/2