In the image below, the larger circle is centered on the origin (0,0). The two circles are tangent and of known radius. The blue line is tangent to the larger circle and passes through the point of the perimeter of the smaller circle shown. (It's always at the 6:00 position.)
Given the input angle of the line (50 degrees), How do I calculate the center position of the smaller circle?
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Let $(a,b)$ be the center of the small circle. The equation of tangent line to the large circle is
$$x\cos50^\circ + y\sin50^\circ =20$$
which passes the 6-clock point $(a, b-11)$, i.e.
$$a\cos50^\circ + (b-11)\sin50^\circ =20\tag1$$
The distance from $(a,b)$ to the origin is
$$a^2+b^2= (11+20)^2\tag2$$
Solve the system of equations (1) and (2) to obtain the center of the small circle. Note that there are two solutions, which indicates two possible locations for the small circle.
Hint:
Let the equation of the line be $y=\tan(140°)\ x + c$. Now use the distance formula (the line is $20$ mm apart from the origin) to solve for $c$. Further, assume the smaller circle to be centered at $(h,k)$. Then its equation will be $$(x-h)^2 + (y-k)^2 = 11^2$$ Now, you can find out the point of intersection of the line and this circle. And finally, use the fact that the line joining this intersection point and the center of the smaller circle $(h,k)$ has slope $\infty$.