What is 16x^2+16y^2+32x+8y=47 ?

The solution to 16x^2+16y^2+32x+8y=47 is Circle with (a,b)=(-1,-1/4),r=2

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16x^2+16y^2+32x+8y=47

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

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

Solution steps

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

Rewrite

16 x^{2} + 16 y^{2} + 32 x + 8 y = 47

in the form of the standard circle equation

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

Therefore the circle properties are:

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

y^{2} = - 16 x

\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1

x = - \frac{1}{8} y^{2}

x^{2} + y^{2} = 13

What is 16x^2+16y^2+32x+8y=47 ?
The solution to 16x^2+16y^2+32x+8y=47 is Circle with (a,b)=(-1,-1/4),r=2