What is 16x^2+16y^2+8x+32y=-1 ?

The solution to 16x^2+16y^2+8x+32y=-1 is Circle with (a,b)=(-1/4 ,-1),r=1

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16x^2+16y^2+8x+32y=-1

16 x^{2} + 16 y^{2} + 8 x + 32 y = - 1

(a, b) = (- \frac{1}{4}, - 1), r = 1

Solution steps

16 x^{2} + 16 y^{2} + 8 x + 32 y = - 1

Rewrite

16 x^{2} + 16 y^{2} + 8 x + 32 y = - 1

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{4}, - 1), r = 1

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

(x + 3)^{2} + (y - 6)^{2} = 49

6 4^{2} = a^{2} + b^{2}

4 x^{2} + 4 x + 4 y^{2} - 16 y = 19

x^{2} + y^{2} = 11

What is 16x^2+16y^2+8x+32y=-1 ?
The solution to 16x^2+16y^2+8x+32y=-1 is Circle with (a,b)=(-1/4 ,-1),r=1