What is 4x^2+32x+4y^2+20y=0 ?

The solution to 4x^2+32x+4y^2+20y=0 is Circle with (a,b)=(-4,-5/2),r=(sqrt(89))/2

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4x^2+32x+4y^2+20y=0

4 x^{2} + 32 x + 4 y^{2} + 20 y = 0

(a, b) = (- 4, - \frac{5}{2}), r = \frac{89}{2}

Solution steps

4 x^{2} + 32 x + 4 y^{2} + 20 y = 0

Rewrite

4 x^{2} + 32 x + 4 y^{2} + 20 y = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- \frac{5}{2}))^{2} = (\frac{89}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 4, - \frac{5}{2}), r = \frac{89}{2}

(x - 4)^{2} + (y - 3)^{2} = 5^{2}

z^{2} + 1^{2} + y^{2} = 16

(x + 3)^{2} + (y + 4)^{2} = 4

x^{2} + y^{2} - 10 x - 4 y + 25 = 0

(x + 3)^{2} + (y + 4)^{2} = 1

What is 4x^2+32x+4y^2+20y=0 ?
The solution to 4x^2+32x+4y^2+20y=0 is Circle with (a,b)=(-4,-5/2),r=(sqrt(89))/2