1
$\begingroup$

I have a crescent defined by two offset circles with different radii: a small one (let's call it outer circle) centered at (0,0) with radius r and a larger one (inner circle) centered at (0,-d) with radius R. The circles are positioned such that they create a lune (rather than a lens) and are separated on the y-axis by distance d. If I draw a ray out from (0,0) to some point on outer circle between its intersection points with inner circle (that is, on the outside edge of the lune), how can I find the point on that ray where it intersects with the perimeter of inner circle?

I need to sort this out because I'm trying to create a function in Python that will return random points within a crescent-shaped field by arbitrarily choosing an angle off of (0,0) toward a point in the outer arc, then picking a valid distance along that arc within the lune. I've worked out a way to select that angle, but I need to figure out what the minimum distance is along that random angle (which is, the point on inner circle's perimeter the ray crosses it at). I know that the range on the radical line (x = 0) is [(R-d)...r], and that there is only one valid point for the cusps where the circles cross, but that isn't super useful to me mathematically.

I've worked out a number of bits of information I need for it using this resource, but I can't quite get my head around this specific problem. I've taken a look around, but I'm not a mathematician and my nomenclature is probably a bit off...

$\endgroup$
  • $\begingroup$ As a footnote, it's actually not necessary that inner circle be larger than outer circle, as long as it intersects with outer circle's perimeter on two points, but I feel like the constraint might simplify things? Maybe not... $\endgroup$ – Augusta Apr 15 '15 at 23:45
0
$\begingroup$

So you have a ray emanating from $(0,0)$ towards a point $(x,y)$. In your statement you assume that $x^2+y^2=r^2$, but that's not a requirement. Any point on that ray can be described as $(\lambda x,\lambda y)$ for some $\lambda>0$. The equation of the inner circle is $x^2+(y+d)^2=R^2$. Plug the point on the ray into that equation, and you get

\begin{align*} (\lambda x)^2+(\lambda y+d)^2&=R^2 \\ (x^2+y^2)\lambda^2 + (2yd)\lambda + (d^2-R^2) &= 0 \end{align*}

which is a quadratic equation in $\lambda$. It has two possible solutions. In the situation you describe, one should be negative, the other positive. You'd want the positive one, since it lies on the ray.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.