What is x(x+5)+(y+2)^2=0 ?

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

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x(x+5)+(y+2)^2=0

x (x + 5) + (y + 2)^{2} = 0

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

Solution steps

x (x + 5) + (y + 2)^{2} = 0

Rewrite

x (x + 5) + (y + 2)^{2} = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 2 x + 6 y - 111 = 0

36 y^{2} + 36 z^{2} = 9

(x - 2)^{2} + (y - 1)^{2} = 81

x^{2} + y^{2} + 4 x - 6 y - 30 = 0

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

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