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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - x + 3 y - \frac{3}{2} = 0

(x + 5)^{2} + (y - 2)^{2} = 11

(x - 2)^{2} + (y + 4)^{2} = 13.19 9^{2}

x^{2} + y^{2} - 16 x - 36 = 0

(x - 1)^{2} + (y + 1)^{2} = 16

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