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 + 10)^{2} + (y + 9)^{2} = 36

x^{2} + 3 x + y^{2} + 7 y - 1 = 0

(x - 9)^{2} + (y - 8)^{2} = 64

(x + \frac{1}{2})^{2} + (y - 5)^{2} = \frac{25}{4}

x^{2} + y^{2} - 4 x + 11 - 84 = 0

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