What is the radius (x+2)^2+(y-3)^2=(-6)^2 ?

The radius (x+2)^2+(y-3)^2=(-6)^2 is r=6

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radius (x+2)^2+(y-3)^2=(-6)^2

radius

(x + 2)^{2} + (y - 3)^{2} = (- 6)^{2}

r = 6

Solution steps

(x + 2)^{2} + (y - 3)^{2} = (- 6)^{2}

Rewrite

(x + 2)^{2} + (y - 3)^{2} = (- 6)^{2}

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 3)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 6

And the radius is:

r = 6

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

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

What is the radius (x+2)^2+(y-3)^2=(-6)^2 ?
The radius (x+2)^2+(y-3)^2=(-6)^2 is r=6