Question 1165331: Find the equation of a circle tangent to the line 3x-4y-16=0 and containing the points (1,8) and (-2,-1).
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
There is surely a better way to find the answer than this....
But it works, so here it is.
Perhaps another tutor will find a simpler but more elegant path to the answer and show it to us.
The calculations required in solving the problem by this method are ugly. I will not show the detailed calculations -- only the results. If you want to learn something from this problem, you will go through the calculations in detail by yourself.
Let the center of the circle be O(h,k).
(1) The center of the circle is on the perpendicular bisector of the segment defined by the two given points, A(1,8) and B(-2,-1).
-- find the slope and midpoint of segment AB
-- find the equation of the perpendicular bisector of AB
That equation is 
Since (h,k) is on that perpendicular bisector, we know  .
(2) Since the circle is tangent to the line 3x-4y-16=0, use the point-to-line distance formula to find that the distance from (h,k) to that line is
(3) The distances from (h,k) to the line and from (h,k) to either of the given points are the same. Use  (from step (1) above) in the formulas for those two distances to get an equation to solve for h, the x-coordinate of the center of the circle.
(The calculations there are REALLY ugly -- use a graphing calculator to solve the equation that says those distances are equal.)
The result of all that ugly calculation is that h=1; that makes k=3.
So the center of the circle is (1,3).
It is then a simple matter (thankfully!) to show that the distances from (1,3) to each of the given points and to the given line are all 5.
And then, finally, the equation of the circle is
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