Hello, novadragon849!

Here are a few of them . . .

You're expected to know the general equation of the circle.

. . The circle with center C(h,k) and radius r has the equation:

. . . . (x - h)² + (y -k)² .= .r²

Do you know how to use that general equation?1) Find the coordinates of the center and the radius of the circle:

. . . . . (x - 7)² + (y + 2)² .= .64

The center is (7,-2) and the radius is 8.

The center is the midpoint of diameter PQ.2) The points P(10,-4) and Q(2,2) are the ends of a diameter.

. . Find the equation of the circle.

We have: .x = ½(10 + 2) = 6 .and .y = ½(-4 + 2) = -1

. . Hence, the center is: C(6,-1)

The radius is the distance CP or CQ.

. . CQ .= .√[(2 - 6)² + (2 + 1)²] .= .√[16 + 9] .= .√[25] .= .5

Therefore, the equation is: .(x - 6)² + (y + 1)² .= .25

3) The point (-2,3) lies on the circle: .(x - 4)² + (y + 1)² .= .r²

. . Find the exact value ofr.

If (-2,3) lies on the circle, then x = -2, y = 3 satisfies the equation.

We have: .(-2 -4)² + (3 + 1)² .= .r² . → . r² .= .(-6)² + 4² .= .36 + 16 .= .52

. . . . . . - . - . . . .__ . . . . .__

Therefore: .r .= .√52 .= .2√13