Question 1165331
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There is surely a better way to find the answer than this....<br>
But it works, so here it is.<br>
Perhaps another tutor will find a simpler but more elegant path to the answer and show it to us.<br>
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.<br>
Let the center of the circle be O(h,k).<br>
(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).<br>
-- find the slope and midpoint of segment AB
-- find the equation of the perpendicular bisector of AB<br>
That equation is {{{y = (-1/3)x+(10/3)}}}<br>
Since (h,k) is on that perpendicular bisector, we know {{{k = (-1/3)h+(10/3)}}}.<br>
(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<br>
{{{abs(3h-4k-16)/sqrt(3^2+4^2) = abs(3h-4k-16)/5}}}<br>
(3) The distances from (h,k) to the line and from (h,k) to either of the given points are the same.  Use {{{k = (-1/3)h-(10/3)}}} (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.<br>
(The calculations there are REALLY ugly -- use a graphing calculator to solve the equation that says those distances are equal.)<br>
The result of all that ugly calculation is that h=1; that makes k=3.<br>
So the center of the circle is (1,3).<br>
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.<br>
And then, finally, the equation of the circle is<br>
{{{(x-1)^2+(y-3)^2 = 25}}}<br>