Question

Find the center of a circle with the equation: x2+y2−4x+2y−11=0

Answer 1

The standard form for the equation of a circle is

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

The center of such a circle is the point with coordinates (h, k).

To get your equation into that form, you have to complete the square (twice; once for x and once for y).

Add 11 to both sides so the constant appears on the right. Build in some space to add a couple of numbers, like this:

$x^2-4x+()+y^2+2y+()=11$

To complete the square on x, look at the coefficient of x (that's -4), take half of it (that's -2), then square it (that's 4).  Add 4 inside the first pair of ().  Do the same kind of thing for y -- half of 2 is 1 and 1 squared is 1;  add 1 inside the second pair of ().  Be sure to add 4 and 1 to the right side, too.

$x^2-4x+4+y^2+2y+1=16$

The first three terms factor and the next 3 terms factor.

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

Match this up with the standard form

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

The center of the circle is (2, -1).

Answer 2

The center of a circle with the equation given is (2,-1).