Find the center and radius of the circle represented by the equation below. x² + y² – 8x – 12y – 48 = 0 Center: (____, ____) Radius: ____

To find the center and radius of the circle represented by the equation (x² + y² – 8x – 12y – 48 = 0), we need to rewrite it in the standard form of a circle’s equation, which is ((x-h)² + (y-k)² = r²), where ((h, k)) is the center and (r) is the radius.

1. Rearrange the equation:

[

x² – 8x + y² – 12y = 48

]

2. Complete the square for (x) and (y):

– For (x² – 8x):

– Half of -8 is -4, and ((-4)² = 16). So, add and subtract 16.
– For (y² – 12y):

– Half of -12 is -6, and ((-6)² = 36). So, add and subtract 36.

The equation becomes:

[

(x² – 8x + 16) + (y² – 12y + 36) = 48 + 16 + 36

]

3. Simplify:

[

(x – 4)² + (y – 6)² = 100

]