If P = (6,5) and Q = (2, 1) are the
endpoints of the diameter of a circle,
find the equation of the circle.
(x – 4)2 + (y - 3)2 = [?]

  (x – 4)^2 + (y – 3)^2 = 8

Step-by-step explanation:

The square of the diameter will be 4 times the square of the radius. The square of the diameter is the sum of squares of differences of coordinates of P and Q:

  d^2 = |PQ|^2 = (2 -6)^2 +(1 -5)^2 = 16 +16

  d^2 = 32

  4r^2 = 32 . . . . . substitute 4r^2 for d^2

  r^2 = 8 . . . . . . . divide by 4 to find r^2

The equation of a circle centered on (h, k) with radius r is ...

  (x -h)^2 +(y -k)^2 = r^2

The given equation tells us the center is (h, k) = (4, 3), so we need only fill in the value of r^2.

  (x -4)^2 +(y -3)^2 = 8

the solution is h=10 because you add 9 to each side of the equation.