Question 256775
If I did this correctly, then the center of the circle should be (x,y) = (2,-1).


you wind up with the equation of:


(x-2)^2 + (y+1)^2 = 16


to graph this equation, you need to solve for y.


subtract (x-2)^2 from both sides of the equation to get:


(y+1)^2 = 16 - (x-2)^2


take square root of both sides of this equation to get:


y+1 = +/- sqrt(11-(x-2)^2)


subtract 1 from both sides of this equation to get:


y = -1 +/- sqrt(16-(x-2)^2)


graph this equation to get:


{{{graph (600,600,-10,10,-10,10,(-1) + sqrt(16-(x-2)^2),(-1) - sqrt(16-(x-2)^2))}}}


from the graph you can see that the center of the circle appears to be x,y = 2,-1.


if you freeze y at -1, then the horizontal range is from x = -2 to x = 6 with x = 2 right in the middle (+/- 4 each way).


if you freeze x at 2, then the vertical range is from y = -5 to y = 3 with y = -1 right in the middle (+/- 4 each way).


the center of the circle is definitely at x,y = (2,-1).


here's how it was derived:


your original equation is:


x^2 + y^2 - 4x + 2y - 11 = 0


reorder the terms to that the x's and the y's are together to get:


x^2 - 4x + y^2 + 2y - 11 = 0


complete the squares on x^2 - 4x to get (x-2)^2 - 4


complete the squares on y^2 + 2y to get (y+1)^2 - 1


your equation becomes:


(x-2)^2 - 4 + (y+1)^2 - 1 - 11 = 0


combine like terms to get:


(x-2)^2 + (y+1)^2 - 16 = 0


add 16 to both sides of the equation to get:


(x-2)^2 + (y+1)^2 = 16


you have just converted to the standard form of the equation for the circle.


that standard form is:


(x-h)^2 + (y-k)^2 = r^2


h is the x coordinate of the center of the circle.
k is the y coodinate of the center of the circle.
r is the radius of the circle.


the answer to problem is:


the center of the circle is at (x,y) = (2,-1).