The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. what is the general form of the equation for the circle? a. x2 + y2 − 6x − 16y + 48 = 0 b. x2 + y2 − 6x − 16y − 25 = 0 c. x2 + y2 + 6x + 16y + 48 = 0 d. x2 + y2 + 6x + 16y − 25 = 0

  A.  x² +y² -6x -16y +48 = 0

Step-by-step explanation:

The standard-form equation for a circle centered at (h, k) with radius r is ...

   (x -h)² +(y -k)² = r²

For your circle, this is ...

  (x -3)² +(y -8)² = 5²

To put this in general form, you subtract the constant on the right, and eliminate parentheses:

  x² -6x +9 +y² -16x +64 -25 = 0

  x² +y² -6x -16y +48 = 0 . . . . . rearrange to descending powers of x, y

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

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step-by-step explanation: