What is (x+3)^2+(y+5)^2=121 ?

The solution to (x+3)^2+(y+5)^2=121 is Circle with (a,b)=(-3,-5),r=11

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(x+3)^2+(y+5)^2=121

(x + 3)^{2} + (y + 5)^{2} = 121

(a, b) = (- 3, - 5), r = 11

Solution steps

(x + 3)^{2} + (y + 5)^{2} = 121

Rewrite

(x + 3)^{2} + (y + 5)^{2} = 121

in the form of the standard circle equation

(x - (- 3))^{2} + (y - (- 5))^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (- 3, - 5), r = 11

What is (x+3)^2+(y+5)^2=121 ?
The solution to (x+3)^2+(y+5)^2=121 is Circle with (a,b)=(-3,-5),r=11