What is (x-11)^2+(y-8)^2=268 ?

The solution to (x-11)^2+(y-8)^2=268 is Circle with (a,b)=(11,8),r=sqrt(268)

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(x-11)^2+(y-8)^2=268

(x - 11)^{2} + (y - 8)^{2} = 268

(a, b) = (11, 8), r = 268

Solution steps

(x - 11)^{2} + (y - 8)^{2} = 268

Rewrite

(x - 11)^{2} + (y - 8)^{2} = 268

in the form of the standard circle equation

(x - 11)^{2} + (y - 8)^{2} = (268)^{2}

Therefore the circle properties are:

(a, b) = (11, 8), r = 268

4 x^{2} - 24 x + 4 y^{2} + 20 y = 39

x^{2} + (y + 1)^{2} = 64

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

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

x^{2} + y^{2} - 4 x + 2 y - 1 = 0

What is (x-11)^2+(y-8)^2=268 ?
The solution to (x-11)^2+(y-8)^2=268 is Circle with (a,b)=(11,8),r=sqrt(268)