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^{2}

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

x^{2} + y^{2} - 10 + 4 y - 7 = 0

(x - 7)^{2} + (y + 2)^{2} = 25

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