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

The radius 16x^2+16y^2+8x-32y=-1 is r=1

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

radius

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

And the radius is:

r = 1

x^{2} + y^{2} - 2 x + 6 y - 15 = 0

2 x^{2} + 2 y^{2} - 14 x + 18 y = 7

x^{2} = 16 - y^{2}

x^{2} + y^{2} = 100

2 x^{2} - 6 x + 2 y^{2} + 2 y = 45

What is the radius 16x^2+16y^2+8x-32y=-1 ?
The radius 16x^2+16y^2+8x-32y=-1 is r=1