There's definitely a solution, but I don't know if there's an elegant way to get to it...
Here's a slightly modified diagram:
I replaced [MATH]X[/MATH] with [MATH]h[/MATH] and [MATH]k[/MATH], and added points [MATH]P1[/MATH] and [MATH]P3[/MATH]. Ultimately, we want to solve for [MATH]k[/MATH].
Recall the standard form of the equation of a circle:
[MATH](x - h)^2 + (y - k)^2 = r^2[/MATH]
The outer circle's center is at [MATH](0, 0)[/MATH] and its radius is [MATH]15[/MATH]. Its equation is [MATH]x^2 + y^2 = 225[/MATH].
From the center of the inner circle, the angle to [MATH]P1[/MATH] is [MATH]180^\circ + 10^\circ[/MATH] and the distance is [MATH]10 + L1[/MATH]. The angle to [MATH]P3[/MATH] is [MATH]-10^\circ[/MATH] and its distance is [MATH]10 + L3[/MATH]. The exact coordinates of [MATH]P1[/MATH] and [MATH]P3[/MATH] are as follows:
- [MATH]P1_x = h + cos(180^\circ + 10^\circ) * (10 + L1)[/MATH]
- [MATH]P1_y = k + sin(180^\circ + 10^\circ) * (10 + L1)[/MATH]
- [MATH]P3_x = h + cos(-10^\circ) * (10 + L3)[/MATH]
- [MATH]P3_y = k + sin(-10^\circ) * (10 + L3)[/MATH]
[MATH]P1[/MATH] and [MATH]P3[/MATH] are on the outer circle. Substituting their coordinates into the outer circle's equation gives us:
- [MATH](h + cos(180^\circ + 10^\circ) * (10 + L1))^2 + (k + sin(180^\circ + 10^\circ) * (10 + L1))^2 = 225[/MATH]
- [MATH](h + cos(-10^\circ) * (10 + L3))^2 + (k + sin(-10^\circ) * (10 + L3))^2 = 225[/MATH]
This is where the elegance falls apart. With this system of equations, we can do some laborious algebra to solve for [MATH]h[/MATH] in terms of [MATH]k[/MATH], then substitute that for [MATH]h[/MATH] in one of the equations to solve for [MATH]k[/MATH] by itself. Since we only have two points on this outer circle, there will be two valid solutions for [MATH]k[/MATH]: one positive and one negative. Since the original problem has the center of the inner circle below the X axis, it's the negative solution we're after.
Wolfram|Alpha
computes this solution as [MATH]k \approx -3.50012[/MATH], which is within the margin of rounding error from what's depicted in the diagram. There may be a simpler way to reach the answer that doesn't involve directly computing sines and cosines, but this method definitely works.
