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I'm trying to figure out a problem that goes like this:

A particle originally placed at the origin tries to reach the point $(12,16)$ whilst covering the shortest distance possible. But there is a circle of radius $3$, centered at the point $(6,8)$, and the point cannot go through the circle. (Click on image to view larger picture.)

Simple Obstacle

My original thought was to travel in a straight line until reaching the circle, and then travel along the circumference until we reach the point on the circumference that is the shortest distance to $(12,16)$. However I feel like this path should be longer than a path along a curve that is tangent to the circle and passes through both the origin and the given point. Now I'm just stuck on how to find this specific curve.

Since the curve must be tangent to the circle at some point I can equate the derivative at some point, but what point exactly?

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    $\begingroup$ This is a nice "toy" example of what are called obstacle problems. As you seem to have intuited, a necessary condition on the path is that at each point along the way, the path either stays in contact with the circle or goes in a straight segment. Here this reduces consideration to travelling around one side of the circle or the other to find a shortest path. $\endgroup$
    – hardmath
    Dec 21, 2019 at 17:19
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    $\begingroup$ For the problem of finding the (two) tangents to a circle from a point (outside the circle) using analytical geometry, see the older Question Tangent to circle. The OP there basically gets the set up correctly but runs into the one possible difficulty, a vertical tangent (where slope $m$ isn't defined). But you will not have this difficulty. $\endgroup$
    – hardmath
    Dec 21, 2019 at 17:46
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    $\begingroup$ Go along the tangent (from origin) till circumference and then along the circumference for a while and then finally along another tangent (but from the red point) till the red point. $\endgroup$ Dec 22, 2019 at 1:39

2 Answers 2

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Here is one way of seeing the shortest path. If you take a rope and try to pull on either end until it is tight. The rope Will show you the shortest path. The rope wont have any angle (sharp corner) on it.

As it had been said in the comments, it will follow a tangent to the circle, then it will wrap around the circle until the tangent that goes thru $(12,16)$.

To evaluate its length, first note that the center of the circle $(6,8)$ is at the middle of the straight line joining $(0,0)$ and $(12,16)$. It means the length from beginning to the circle is the same as the one from the circle to the end.

Second, we know that a tangent is perpendicular to the radius. We have a right triangle formes by the origin $O(0,0)$, the center of the circle $C(6,8)$ and the point where the tangent meet the circle $P$. In the triangle $OCP$, we know that $P$ is a right angle, $PC =3$ is the radius of the circle and $OC=10$. Then $$OP=\sqrt{10^2-3^2}=\sqrt{91}$$ We now have to find the length of rope that wrap around the circle. The angle it follow the circle is $$\pi-2*\angle{PCO}$$ $$\pi-2*\arccos\left(\frac3{10}\right)$$ And the length is $$3\pi-6*\arccos\left(\frac3{10}\right)$$ Finally, the shortest path is equal to $$2*\sqrt{91}+3\pi-6*\arccos\left(\frac3{10}\right)=20.906\dots$$ I can't add a picture with my phone. I'll add one as soon as I can.

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Hint: (Developing a little more the comment by Harshal Gajjar)

You have two rectangular triangles, and you could know the position of the vertex, where the angle A is, since we know that the sides of this triangle are $3,\sqrt 91$ and $10$.

$\qquad\qquad\qquad\qquad\qquad$ enter image description here

My calculations give me that this vertex has coordinates $\left(6+3\frac{9-\sqrt 91}{50},8+3\frac{\sqrt{2328}}{50}\right)$.

Since the positions $(0,0)$ and $(12,16)$ are symmetrically allocated with respect to the center of the circle, the calculations for the triangle based on the origin is similar.

The final step is the calculation of the little arc on the circle between the points of tangency.

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