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

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

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

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

(a, b) = (- 4, - 2), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, - 2), r = 2

x, x^{2} + Y (x)^{2} = 9

3 x^{2} + 3 y^{2} + 6 y - 20 = 0

x^{2} + y^{2} > 9

x^{2} + y^{2} - 4 x + 2 y + 1 = 0

(x - 6)^{2} + (y + 36)^{2} = 25

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