What is (x+5)^2+(y-8)^2=121 ?

The solution to (x+5)^2+(y-8)^2=121 is Circle with (a,b)=(-5,8),r=11

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(x+5)^2+(y-8)^2=121

(x + 5)^{2} + (y - 8)^{2} = 121

(a, b) = (- 5, 8), r = 11

Solution steps

(x + 5)^{2} + (y - 8)^{2} = 121

Rewrite

(x + 5)^{2} + (y - 8)^{2} = 121

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 8)^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (- 5, 8), r = 11

5 x^{2} + 5 y^{2} - 7 x + 11 y - 9 = 0

x^{2} + 10 x + y^{2} - 12 y = 3

x^{2} + y^{2} - 22 y - 5 = 0

x^{2} + y^{2} + 6 x + 8 y - 75 = 0

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} = 16

What is (x+5)^2+(y-8)^2=121 ?
The solution to (x+5)^2+(y-8)^2=121 is Circle with (a,b)=(-5,8),r=11