What is y^2=-(x+2)(x-3) ?

The solution to y^2=-(x+2)(x-3) is Circle with (a,b)=(1/2 ,0),r= 5/2

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y^2=-(x+2)(x-3)

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

(a, b) = (\frac{1}{2}, 0), r = \frac{5}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - 0)^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 0), r = \frac{5}{2}

(x - 8)^{2} + (y - 9)^{2} = 3

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

(x + 2)^{2} + y^{2} = 10

x^{2} + z^{2} = 5

x^{2} + z^{2} = 2

What is y^2=-(x+2)(x-3) ?
The solution to y^2=-(x+2)(x-3) is Circle with (a,b)=(1/2 ,0),r= 5/2