Question

The equation of a circle is x^2 + y^2 -4x + 2y - 11 = 0. What are the center and the radius of the circle? Show your work.

Answer 1

Center at (-2, -1) and radius = 4

Explanation:

We are given the following equation of a circle which we want to convert into the standard form:

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

Note that

(x + a)² = x² + 2ax + a²

so that

x² + 2ax = (x + a)² - a²

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

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

`(x-2)^2-2^2+(y+1)^2+1^2=11`

`(x-2)^2+(y+1)^2-5=11`

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

Therefore, this equation represents a circle with center at (-2, -1) and radius √16 = 4, as shown below.

Answer 2

The center of the circle;

(2, -1)

The radius of the circle;

r = 4

Explanation:

The first step is to complete the square on both x and y;

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

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

We determine c1 to complete the square on x;

c1 = (b/2)^2

c1 = (-4/2)^2 = 4

We then determine c2 to complete the square on y;

c2 = (2/2)^2 = 1

The equation of the circle is then re-written as;

x^2 -4x + 4 +y^2 +2y + 1 = 11 +4 +1

We then factorize the expressions in x and y separately;

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

The center of the circle is thus;

(2, -1)

The radius is the square root of 16;

r = 4

find the attached.