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 - 4)^{2} - 9 = 0

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

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

(x - 5.5)^{2} + (y - 5.5)^{2} - (1.5)^{2} = 0

x^{2} + y^{2} - 10 x + 2 y + 10 = 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