Question 551035: Find the equation of a circle tangant to the axes and passing through (2,1).
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Find the equation of a circle tangant to the axes and passing through (2,1)
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The equation for a circle with center (a,b) and radius R is (x-a)^2 + (y-b)^2 = R^2
If the circle is tangent to both axes, then the points of tangency are (R,0) and (0,R)
Therefore, the circle must be centered at the point (R,R)
The circle passes through the point (2,1):
(2-R)^2 + (1-R)^2 = R^2
Solve for R:
4 - 4R + R^2 + 1 - 2R + R^2 = R^2
R^2 - 6R + 5 = 0
(R-5)(R-1) = 0
This gives two solutions, R=1 and R=5
So there are two circles which satisfy the requirements:
(x-1)^2 + (y-1)^2 = 1
(x-5)^2 + (y-5)^2 = 25
The circles are graphed below. Note that both circles pass through the point (2,1)
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