What is the center of the circle and the radius

X2+y2-4x+12y-24=0

Subtract

31

31

from both sides of the equation.

x

2

+

y

2

-

4

x

-

12

y

=

-

31

x2+y2-4x-12y=-31

Complete the square for

x

2

-

4

x

x2-4x

.

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(

x

-

2

)

2

-

4

(x-2) 2-4

Substitute

(

x

-

2

)

2

-

4

(x-2) 2-4

for

x

2

-

4

x

x2-4x

in the equation

x

2

+

y

2

-

4

x

-

12

y

=

-

31

x2+y2-4x-12y=-31

.

(

x

-

2

)

2

-

4

+

y

2

-

12

y

=

-

31

(x-2) 2-4+y2-12y=-31

Move

-

4

-4

to the right side of the equation by adding

4

4

to both sides.

(

x

-

2

)

2

+

y

2

-

12

y

=

-

31

+

4

(x-2) 2+y2-12y=-31+4

Complete the square for

y

2

-

12

y

y2-12y

.

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(

y

-

6

)

2

-

36

(y-6) 2-36

Substitute

(

y

-

6

)

2

-

36

(y-6) 2-36

for

y

2

-

12

y

y2-12y

in the equation

x

2

+

y

2

-

4

x

-

12

y

=

-

31

x2+y2-4x-12y=-31

.

(

x

-

2

)

2

+

(

y

-

6

)

2

-

36

=

-

31

+

4

(x-2) 2 + (y-6) 2-36=-31+4

Move

-

36

-36

to the right side of the equation by adding

36

36

to both sides.

(

x

-

2

)

2

+

(

y

-

6

)

2

=

-

31

+

4

+

36

(x-2) 2 + (y-6) 2=-31+4+36

Simplify

-

31

+

4

+

36

-31+4+36

.

(

x

-

2

)

2

+

(

y

-

6

)

2

=

9

(x-2) 2 + (y-6) 2=9

This is the form of a circle. Use this form to determine the center and radius of the circle.

(

x

-

h

)

2

+

(

y

-

k

)

2

=

r

2

(x-h) 2 + (y-k) 2=r2

Match the values in this circle to those of the standard form. The variable

r

r

represents the radius of the circle,

h

h

represents the x-offset from the origin, and

k

k

represents the y-offset from origin.

r

=

3

r=3

h

=

2

h=2

k

=

6

k=6

The center of the circle is found at

(

h

,

k

)

(h, k)

.

Center:

(

2

,

6

)

(2,6)

These values represent the important values for graphing and analyzing a circle.

Center:

(

2

,

6

)

(2,6)

Radius:

3

3