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

The solution to 16x^2+16y^2+96x-40y-315=0 is Circle with (a,b)=(-3, 5/4),r= 11/2

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

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

(a, b) = (- 3, \frac{5}{4}), r = \frac{11}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{5}{4})^{2} = (\frac{11}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{5}{4}), r = \frac{11}{2}

x^{2} + y^{2} - 4 x + 2 y - 20 = 0

(x - 1)^{2} + (y + 4)^{2} = 9

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

(x - 1)^{2} + (y - 3)^{2} = 4

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

What is 16x^2+16y^2+96x-40y-315=0 ?
The solution to 16x^2+16y^2+96x-40y-315=0 is Circle with (a,b)=(-3, 5/4),r= 11/2