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= 3/(2sqrt(2))

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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 = \frac{3}{2 2}

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} = (\frac{3}{2 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{4}, 1), r = \frac{3}{2 2}

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{25} = 1

x^{2} + y^{2} + 9 y - 2 = 0

x^{2} + y^{2} - 2 x - 4 y - 15 = 0

4 x^{2} + 4 y^{2} - 24 y + 51 = 16 x

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

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= 3/(2sqrt(2))