What is 3^2=(x-0)^2+(y+2)^2 ?

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

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3^2=(x-0)^2+(y+2)^2

3^{2} = (x - 0)^{2} + (y + 2)^{2}

(a, b) = (0, - 2), r = 3

Solution steps

3^{2} = (x - 0)^{2} + (y + 2)^{2}

Rewrite

3^{2} = (x - 0)^{2} + (y + 2)^{2}

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (0, - 2), r = 3

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

9 = x^{2} - 6 x + 9 + y^{2} + 4 y + 4

4 x^{2} + 4 y^{2} = 256

3 x^{2} + 3 y^{2} - 6 x + 3 y - 4 = 0

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

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