What is 4x^2+4y^2+16x+24y+31=0 ?

The solution to 4x^2+4y^2+16x+24y+31=0 is Circle with (a,b)=(-2,-3),r=(sqrt(21))/2

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4x^2+4y^2+16x+24y+31=0

4 x^{2} + 4 y^{2} + 16 x + 24 y + 31 = 0

(a, b) = (- 2, - 3), r = \frac{21}{2}

Solution steps

4 x^{2} + 4 y^{2} + 16 x + 24 y + 31 = 0

Rewrite

4 x^{2} + 4 y^{2} + 16 x + 24 y + 31 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 3))^{2} = (\frac{21}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 3), r = \frac{21}{2}

x^{2} + y^{2} - 4 x + y - 4 = 9

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

1 (x^{2} + y^{2}) = x

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

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

What is 4x^2+4y^2+16x+24y+31=0 ?
The solution to 4x^2+4y^2+16x+24y+31=0 is Circle with (a,b)=(-2,-3),r=(sqrt(21))/2