Part a.
r = 10 is the radius
(h,k) = (-9,-2) is the center
The general circle equation is
(x-h)^2 + (y-k)^2 = r^2
Plugging the given values into that equation gives us this
(x-(-9))^2 + (y-(-2))^2 = 10^2
(x+9)^2 + (y+2)^2 = 100
Answer: (x+9)^2 + (y+2)^2 = 100
=================================
Part b.
Four easy points we can find are the points that are aligned on the north, east, south and west parts of the circle.
Start at (-9,-2) which is the center of the circle. Move up 10 units and you'll get to (-9,8) which is the point at the northern most part of the circle. We move 10 units because this is the radius.
Go back to the center (-9,-2). Move 10 units to the right and you'll arrive at (1,-2)
Go back to the center again and move 10 units down to get to (-9,-12)
Finally, go back to the center and move 10 units to the left to arrive at (-19,-2)
Answers: (-9,8) and (1,-2) and (-9,-12) and (-19,-2)
=================================
Part c.
The point (-3,6) has x = -3 and y = 6 pairing up.
Plug these values into the equation we found back in part a.
Then simplify both sides. If we get the same number on each side, then we will have a true equation and that will confirm (-3,6) is on the circle. Otherwise, we get a false equation and the point is not on the circle.
-------
(x+9)^2 + (y+2)^2 = 100
(-3+9)^2 + (6+2)^2 = 100
(6)^2 + (8)^2 = 100
36 + 64 = 100
100 = 100
We get the same thing on both sides, so we have a true equation and (-3,6) is on the circle.
Answer: Yes, the point (-3,6) is on the circle.
-------
Below is the graph showing this point on the circle along with the other four points found earlier back in part b.