What is (x+1)^2+(y-2)^2=12 ?

The solution to (x+1)^2+(y-2)^2=12 is Circle with (a,b)=(-1,2),r=sqrt(12)

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(x+1)^2+(y-2)^2=12

(x + 1)^{2} + (y - 2)^{2} = 12

(a, b) = (- 1, 2), r = 12

Solution steps

(x + 1)^{2} + (y - 2)^{2} = 12

Rewrite

(x + 1)^{2} + (y - 2)^{2} = 12

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 2)^{2} = (12)^{2}

Therefore the circle properties are:

(a, b) = (- 1, 2), r = 12

(12)^{2} + 8 = x^{2} + y^{2}

(x + 1)^{2} + (y + 1)^{2} = 6

(x - (- 7))^{2} + (y - (- 4))^{2} = 7^{2}

4 x^{2} + 12 x + 4 y^{2} + 16 y + 11 = 0

(x + 5)^{2} + (y + 6)^{2} = 12

What is (x+1)^2+(y-2)^2=12 ?
The solution to (x+1)^2+(y-2)^2=12 is Circle with (a,b)=(-1,2),r=sqrt(12)