What is 8x^2+8y^2-32x+48y-16=0 ?

The solution to 8x^2+8y^2-32x+48y-16=0 is Circle with (a,b)=(2,-3),r=sqrt(15)

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8x^2+8y^2-32x+48y-16=0

8 x^{2} + 8 y^{2} - 32 x + 48 y - 16 = 0

(a, b) = (2, - 3), r = 15

Solution steps

8 x^{2} + 8 y^{2} - 32 x + 48 y - 16 = 0

Rewrite

8 x^{2} + 8 y^{2} - 32 x + 48 y - 16 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = (15)^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = 15

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

3 x^{2} + 3 y^{2} - 48 = 0

x^{2} + y^{2} + 20 x - 69 = 0

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

2 x^{2} + 2 y^{2} - 3 x + 4 y - 1 = 0

What is 8x^2+8y^2-32x+48y-16=0 ?
The solution to 8x^2+8y^2-32x+48y-16=0 is Circle with (a,b)=(2,-3),r=sqrt(15)