Answer: (x-5)^2 + (y-7)^2 = 25
Center = (5,7)
Radius = 5
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Explanation:
The circle must be tangent to the y axis, and it has a radius of 5, so that means the center (h,k) must either be on the line x = 5 or x = -5, since those vertical lines are exactly five units away from the y axis.
But we're also told that (10,7) must be on the circle. That rules out x = -5 because the distance from (10,7) to (-5,k) is going to be larger than 5. I recommend drawing out a right triangle to see why this is the case.
So the center (h,k) must be on the line x = 5.
The center is the point (h,k) = (5,k)
Plug that h value along with (x,y) = (10,7) and r = 5 into the equation below
Solve for k.
(x-h)^2 + (y-k)^2 = r^2
(x-5)^2 + (y-k)^2 = 5^2 .... plug in h = 5 and r = 5
(10-5)^2 + (7-k)^2 = 25 ... plug in x = 10 and y = 7
25 + (7-k)^2 = 25
(7-k)^2 = 25-25
(7-k)^2 = 0
7-k = sqrt(0)
7-k = 0
-k = 0-7
-k = -7
k = 7
The center is (h,k) = (5,7)
We have (x-h)^2 + (y-k)^2 = r^2 updating to (x-5)^2 + (y-7)^2 = 25
