How to adjust spaceship's speed? A spaceship is moving through 2-dimensional space by a series of jumps. Each jump consists of rotating right and flying forward. The angle of rotation is always the same and the distance flown forward is always the same. Thus, ship is flying circles around some point.

There's a special point marked in space. The question is: with given angle, how far should the ship travel with each jump, such that circle he's flying will cross the point? (ship does not necessarily need to hit the point). Is there only one solution?
Ship may turn left or right, but this decision must be made before starting flying, later it always turns the same way each jump.

Result
Just for fun, this is my ship navigation algorithm, that was developed with amd's aid:

 A: Assuming that the initial heading of the ship is upwards (in the positive $y$ direction) and that $\theta$ increases clockwise, let $d$ be the hop distance. W.l.o.g. we can assume that the ship starts at the origin. Its first two hops will take it to $(d\sin\theta,d\cos\theta)$ and $(d\sin\theta+d\sin2\theta,d\cos\theta+d\cos2\theta)$. We now have three points on the implied circle and can immediately write down its equation in the form $$\det\begin{bmatrix}x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{bmatrix} = 0,$$ specifically, $$\det\begin{bmatrix}x^2+y^2 & x & y & 1 \\
0&0&0&1 \\
d^2 & d\sin\theta & d\cos\theta & 1 \\
d^2(\cos\theta+\cos2\theta)^2+d^2(\sin\theta+\sin2\theta)^2 & d(\sin\theta+\sin2\theta) & d(\cos\theta+\cos2\theta) & 1
\end{bmatrix} = 0$$ which simplifies to $$x^2+y^2-\left(d\cot\frac\theta2\right)x+dy = 0.\tag{*}$$ (Subtract the third row from the fourth to simplify the computation of this determinant.) Solving this for $d$ yields $${x^2+y^2 \over x\cot\frac\theta2-y}.$$ If $d\lt0$, i.e., if $x\cot\frac\theta2\lt y$, that means that the circle would have to be traversed backwards: the target point can’t be reached with the given turn rate.  

* This equation is equivalent to $\left(x-\frac12d\cot{\frac\theta2}\right)^2 + \left(y+\frac d2\right)^2 = \left({d \over 2\sin{\frac\theta2}}\right)^2 = {d^2\over 2-2\cos\theta}$, which could be useful if you need to know more about the circle.
