What is the radius (x+4)^2+(y-3)^2=16 ?

The radius (x+4)^2+(y-3)^2=16 is r=4

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radius (x+4)^2+(y-3)^2=16

radius

(x + 4)^{2} + (y - 3)^{2} = 16

r = 4

Solution steps

(x + 4)^{2} + (y - 3)^{2} = 16

Rewrite

(x + 4)^{2} + (y - 3)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, 3), r = 4

And the radius is:

r = 4

x^{2} + y^{2} + 4 x + 14 y + 52 = 0

x^{2} + y^{2} - 14 x + 8 y - 16 = 0

(x - 3)^{2} + (y + 3)^{2} = 36

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

x^{2} + y^{2} + 6 x + 2 y - 54 = 0