Mock AIME II 2012 Problems/Problem 6
Problem
A circle with radius 
and center in the first quadrant is placed so that it is tangent to the 
-axis. If the line passing through the origin that is tangent to the circle has slope 
, then the -coordinate of the center of the circle can be written in the form 
where 
, 
, and 
are positive integers, and 
. Find 
.
![[asy] size(200); real y = (5+sqrt(125))/2; defaultpen(linewidth(0.8)); draw(origin--(0,15),EndArrow); draw(origin--(15,0),EndArrow); draw(circle((5,y),5)); draw(origin--(15,15/2)); label("$\beta$",(3/2,3/8),dir(aTan(1/2)/2)); label("$y=\frac{1}{2}x$",(11,5),S); dot((5,y)); draw(origin--(5,y)--(0,y),linetype("4 4")); pair tan = reflect(origin,(5,y))*(0,y); draw((5,y)--tan,linetype("4 4")); [/asy]](http://latex.artofproblemsolving.com/9/e/5/9e565f692e5033b68a03f322f939bc72e8c70649.png)
Solution 1
Since the circle has a radius of , is tangent to the y-axis, and has its center in the first quadrant, its center has coordinates 
for some positive 
. Also, since the circle is tangent to the line 
, the distance from the center to that line must be . The equation for the line can be rewritten as 
. Using the point-to-line formula, we must have 
. Simplifying this, we have 
. This gives two solutions for . We have 
or 
. We reject the negative value, and so we have , and 
.
Solution 2
Let 
be the angle between the line and the 
-axis. Then we have 
, so 
and 
. Drawing in the line from the center of the circle to the origin, we see that this line bisects the angle between the -axis and the line , so its tangent is equal to 
. Now using the upper right triangle, we see that 
also equals 
, where is the -coordinate of the center of the circle. Thus, we have 
, so 
and 
. Our answer is thus 
.
