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

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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 $sqrt64=8$ .

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