SOLUTION: Find the equation of the circle tangent to x=0; center on x+y=10; contains (2,9)

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Question 834912: Find the equation of the circle tangent to x=0; center on x+y=10; contains (2,9)
Answer by reviewermath(1029) About Me  (Show Source):
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Q:
Find the equation of the circle tangent to x=0; center on x+y=10; contains (2,9)
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A:
Let (h, k) be the center and r the radius of the circle.
The center lies on the line x + y = 10, so h + k = 10 or h = 10 - k.
The distance between the line x = 0 and the center (h, k) is h or 10 - k.
The distance between (h, k) and (2, 9) is equal to
sqrt%28%28h+-+2%29%5E2+%2B+%28k+-+9%29%5E2%29, substitute h = 10 - k
= sqrt%28%28k+-+8%29%5E2+%2B+%28k+-+9%29%5E2%29
The two distances are equal (both are equal to the length of the radius).
sqrt%28%28k+-+8%29%5E2+%2B+%28k+-+9%29%5E2%29 = 10 - k, square both sides
%28k+-+8%29%5E2+%2B+%28k+-+9%29%5E2+=+k%5E2+-+20k+%2B+100
k%5E2+-+14k+%2B+45+=+0
(k - 5)(k - 9) = 0
k = 5 or 9
If k = 5:
then h = 10 - 5 = 5, so the center is (5, 5) and the radius is r = 5.
The equation is %28x+-+5%29%5E2+%2B+%28y+-+5%29%5E2+=+25
If k = 9:
then h = 10 - 9 = 1, so the center is (1, 9) and the radius is r = 1.
The equation is %28x+-+1%29%5E2+%2B+%28y+-+9%29%5E2+=+1
Graph:

Answers: highlight%28%28x+-+5%29%5E2+%2B+%28y+-+5%29%5E2+=+25%29 or highlight%28%28x+-+1%29%5E2+%2B+%28y+-+9%29%5E2+=+1%29