What is the radius (x+12)^2+(y-13)^2-25=0 ?

The radius (x+12)^2+(y-13)^2-25=0 is r=5

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radius (x+12)^2+(y-13)^2-25=0

radius

(x + 12)^{2} + (y - 13)^{2} - 25 = 0

r = 5

Solution steps

(x + 12)^{2} + (y - 13)^{2} - 25 = 0

Rewrite

(x + 12)^{2} + (y - 13)^{2} - 25 = 0

in the form of the standard circle equation

(x - (- 12))^{2} + (y - 13)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 12, 13), r = 5

And the radius is:

r = 5

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

(x + 1)^{2} + y^{2} = 246

5 x^{2} - 10 x + 5 y^{2} + 3 y - 7 = 0

x^{2} + y^{2} + 8 x + 6 y = 0

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{9}{4}

What is the radius (x+12)^2+(y-13)^2-25=0 ?
The radius (x+12)^2+(y-13)^2-25=0 is r=5