Mock AIME II 2012 Problems/Problem 6
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 .
Diagram used for both solutions
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 .
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 .