How to demonstrate a limit without having to pre-calculate certain points

Question

1. You have a circle of radius r centered at the origin.

2. You have a circle of radius 1 centered at (1,0).

3. You draw a ray through the point (0,r) and the point where the circles intersect.

4. Let R be the point where the ray intersects the x-axis.

5. The question is, what happens to the point R as $r\to 0$.

I have:

Manipulate[
  Graphics[{
    Circle[{1, 0}, 1],
    Circle[{0, 0}, r],
    Red, PointSize[Medium],
    Point[{0, r}],
    Point[{r^2/2, Sqrt[1 - (r^2/2 - 1)^2]}],
    Point[{2 + Sqrt[4 - r^2], 0}],
    HalfLine[{{0, r}, {r^2/2, Sqrt[1 - (r^2/2 - 1)^2]}}]},
    Axes -> True,
    PlotRange -> {{-2, 5}, {-2, 2}}], 
  {{r, 1.4}, 0, 1.5}]

Which produces this image.

Now, I did a bunch of math, similar triangles, etc., to find that the coordinates of the point of intersection is $(r^2/2,\sqrt{1-(r^2/2-1)^2}$ and the point R, where the ray intersects the x-axis, is $2+\sqrt{4-r^2}$.

I'm just wondering how folks might do this same thing without making those math calculations.

Answer 1

Here is an example using geometrical region functions:

Manipulate[

 pointoncircle = RegionIntersection[
    Circle[{0, 0}, r],
    Circle[{1, 0}, 1],
    HalfPlane[{{0, 0}, {1, 0}}, {1, 1}]
  ];
 ray = HalfLine[{{0, r}, pointoncircle[[1]]}];
 pointonaxis = RegionIntersection[ray, InfiniteLine[{{0, 0}, {1, 0}}]];

 Graphics[{
   Circle[{0, 0}, r], Circle[{1, 0}, 1],
   Red, PointSize[Large],
   Point[{0, r}], pointoncircle, ray, pointonaxis
  },
  Axes -> True, PlotRange -> {{-2, 5}, {-2, 2}}
 ],

 {{r, 1.4}, 0.001, 1.5}
]

---

Of course an alternative approach would be to set up appropriate equations describing the points you need, then letting Mathematica do the solving:

pointoncircle = Point[{x, y}] /. First@
   NSolve[{x^2 + y^2 == r^2, (x - 1)^2 + y^2 == 1, y >= 0}, {x, y}];

ray = HalfLine[{{0, r}, pointoncircle[[1]]}];

pointonaxis = Point[{x, 0} /. First@
    Solve[InterpolatingPolynomial[{{0, r}, pointoncircle[[1]]}, x] == 0, x]];

These definitions can be plugged into the Manipulate shown above with identical results.