Part A
We have the point (-1,1) so x = -1 and y = 1. Plug these values into the equation to get
(x-3)^2 + y^2 = 49
(-1-3)^2 + (1)^2 = 49
(-4)^2 + (1)^2 = 49
16 + 1 = 49
17 = 49
The equation is false so the point (-1,1) is NOT on the circle edge. The fact that the left side is smaller means that the point is on the interior of the circle
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Part B
Repeat the same steps as part A, but now use (x,y) = (10,0)
(x-3)^2 + y^2 = 49
(10-3)^2 + 0^2 = 49
7^2 + 0^2 = 49
49+0 = 49
49 = 49
This equation is true, so (10,0) is on the circle edge
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Part C
Plug in (x,y) = (4,-8)
(x-3)^2 + y^2 = 49
(4-3)^2 + (-8)^2 = 49
(-1)^2 + (-8)^2 = 49
1 + 64 = 49
65 = 49
The equation is false, so the point is NOT on the circle edge.
The left side is larger so the point on the exterior of the circle.
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Summary:
point A is inside the circle
point B is on the circle's edge
point C is on the outside of the circle
See attached for visual proof