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