What is (x+1)^2+y^2=4 ?

The solution to (x+1)^2+y^2=4 is Circle with (a,b)=(-1,0),r=2

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(x+1)^2+y^2=4

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

(a, b) = (- 1, 0), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (- 1, 0), r = 2

(x - 2)^{2} + (y + 1)^{2} = 25

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

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

(x - 2)^{2} + (y - 1)^{2} = 25

x^{2} + (y - 4)^{2} = 16

What is (x+1)^2+y^2=4 ?
The solution to (x+1)^2+y^2=4 is Circle with (a,b)=(-1,0),r=2