What is the radius (x+1)^2+(y+3)^2=4 ?

The radius (x+1)^2+(y+3)^2=4 is r=2

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radius (x+1)^2+(y+3)^2=4

radius

(x + 1)^{2} + (y + 3)^{2} = 4

r = 2

Solution steps

(x + 1)^{2} + (y + 3)^{2} = 4

Rewrite

(x + 1)^{2} + (y + 3)^{2} = 4

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 2

x^{2} + y^{2} - 64 = 0

3 x^{2} + 5 x + 3 y^{2} + 6 y - 15 = 0

x^{2} + y^{2} + 8 x - 14 y - 56 = 0

x^{2} + y^{2} + 3 x - 5 y + 1 = 0

x^{2} + y^{2} = 1 + 2 y