What is the center and radius of the circle with equation #x^2 + y^2 – 4x + 22y + 61 = 0#?

Question

1521 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-03T22:39:54

The center is at #(2,-11)# and the radius is #8#.

All the manipulations that will be done are in effort to make the equation resemble the general form of a circle:

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

Where the center is #(h,k)# and the radius is #r#.

Sort the #x# and #y# terms and move the constant to the other side.

#x^2-4x+y^2+22y=-61#

Complete the square for the #x# and #y# parts. Balance the equation by adding the constants to the left and right sides.

#(x^2-4x+4)+(y^2+22y+121)=-61+4+121#

#(x-2)^2+(y+11)^2=64#

Thus, the center is at #(2,-11)# and the radius is #sqrt64=8#.