Question 714404: Find the equation of the circle passing through (2,2) and tangent to the line y=1 and y= 6
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The tangent lines touch the circle at the top and the bottom.
This means that the diameter of the circle is the distance between the two tangent lines 6-1=5.
So r = 2.5, and the center of the circle lies halfway in between y=1 and y=6.
So the center has the coordinates (a,3.5) and the radius is 2.5:
(x-a)^2 + (y-3.5)^2 = 2.5^2
We can use the point (2,2) on the circle to find a:
(2-a)^2 + (2-3.5)^2 = 2.5^2
This simplifies to a^2 - 4a = 0
a(a-4) = 0
There are two possibilities, a=0 and a=4.
So there are two circles which fit the criteria:
x^2 + (y-3.5)^2 = 6.25
(x-4)^2 + (y-3.5)^2 = 6.25
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