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

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

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

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

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

Solution steps

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

Rewrite

(x - 2)^{2} + (y - (- 3))^{2} = 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 \cdot 1)^{2} + (y - 1)^{2} = 5

\frac{x ^{2}}{2} + \frac{y ^{2}}{2} = 18

a^{2} + b^{2} = a + 2

x^{2} + y^{2} - 8 x - 6 y - 16 = 0

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

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