What is the radius (x+5)^2+(y-7)^2=289 ?

The radius (x+5)^2+(y-7)^2=289 is r=17

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radius (x+5)^2+(y-7)^2=289

radius

(x + 5)^{2} + (y - 7)^{2} = 289

r = 17

Solution steps

(x + 5)^{2} + (y - 7)^{2} = 289

Rewrite

(x + 5)^{2} + (y - 7)^{2} = 289

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 7)^{2} = 1 7^{2}

Therefore the circle properties are:

(a, b) = (- 5, 7), r = 17

And the radius is:

r = 17

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