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 + 3)^{2} + (y - 3)^{2} = 25

x^{2} + y^{2} + 14 x - 2 y = 225

- x^{2} - y^{2} + 5 = x + y + 1

x^{2} + y^{2} + 8 x - 6 y = 6

x^{2} - 8 x + y^{2} - 2 y - 19 = 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