What is (x-(-2))^2+(y-(-1))^2=(3)^2 ?

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

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(x-(-2))^2+(y-(-1))^2=(3)^2

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

(a, b) = (- 2, - 1), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 1), r = 3

x^{2} + y^{2} - 9 x + 4 y - 138 = 0

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

x^{2} - 6 x + y^{2} + 8 y + 5 = 0

(x - (- 5))^{2} + (y - 0)^{2} = 4^{2}

(x + 1)^{2} + (y - 4)^{2} = 4

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