SOLUTION: Find the equation of the circle which touches the X axis at the point (3,0) and cuts off an intercept 8 units from the positive part of the y axis

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Question 1039233: Find the equation of the circle which touches the X axis at the point (3,0) and cuts off an intercept 8 units from the positive part of the y axis
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of the circle which touches the X axis at the point (3,0) and cuts off an intercept 8 units from the positive part of the y axis
The standard equation of any circle is
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
Since it is tangent to the x-axis at (3,0),
the center is directly above (3,0), so its
center also has x-coordinate 3. So h=3
%28x-3%29%5E2%2B%28y-k%29%5E2=r%5E2
It passes through (x,y) = (3,0), so
%283-3%29%5E2%2B%280-k%29%5E2=r%5E2
0%5E2%2Bk%5E2=r%5E2
k%5E2=r%5E2
k and r are both positive numbers.
k=r and
%28x-3%29%5E2%2B%28y-r%29%5E2=r%5E2
It has a y-intercept at (0,8)
%280-3%29%5E2%2B%288-r%29%5E2=r%5E2
%28-3%29%5E2%2B64-16r%2Br%5E2=r%5E2
9%2B64-16r%2Br%5E2=r%5E2
73-16r=0
-16r=-73
r=%28-73%29%2F%28-16%29
r=73%2F16
And since x=r=73%2F16,
%28x-3%29%5E2%2B%28y-73%2F16%29%5E2=%2873%2F16%29%5E2