What is the radius (x-1)^2+(y+3)^2-9=0 ?

The radius (x-1)^2+(y+3)^2-9=0 is r=3

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radius (x-1)^2+(y+3)^2-9=0

radius

(x - 1)^{2} + (y + 3)^{2} - 9 = 0

r = 3

Solution steps

(x - 1)^{2} + (y + 3)^{2} - 9 = 0

Rewrite

(x - 1)^{2} + (y + 3)^{2} - 9 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the radius is:

r = 3

(x + 2)^{2} + (y - 4)^{2} = 9

x^{2} + y^{2} - 4 x - 6 y - 23 = 0

(x - 1)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} - 2 y - 15 = 0

x^{2} + y^{2} + 2 x = 3 y