Shortest path to a point that doesn't pass through the given circle

[JDStupi]

Homework Statement

I am trying to work through a problem solving book, and one of the problems is to try to find the shortest path leading from (0,0) to (12,16) that doesn't pass through the circle (x-6)^2+(y-8)^2=25

Relevant Equations

(x-6)^2+(y-8)^2=25
and the various line formulas.

This is my attempt at a solution. Point A is the center of the circle (6,8) and Point B is the given point (12,16). I believe that the shortest path would be the one that is equal to the sum of CE and EB or its symmetrical complement. (I forgot to put a point where the top line intersects the y-axis). My problem is 1) To show that this is the case and 2) To find an analytical expression for this.

I must find the slope of a line that makes it from the origin to the point E = (12,y) and lies tangent to the circle at point D. From visual inspection, this point lies to the right of the minimum height of the circle, but how do I determine this optimal point, when I do not know the optimal y for point E?

From there, the solution is a straightforward sum.

I'm not looking for the exact answer, I want coaching on the thinking involved and the process. Thank you all for your help.

 

[PeroK]

I'm not looking for the exact answer, I want coaching on the thinking involved and the process. Thank you all for your help.

Hint 1: circles and lines have equations that define them

Hint 2: a tangent to a circle forms a right-angle with the radius of the circle at that point.

Although in your diagram CDA doesn't look like a right-angle triangle it is!

 

[LCKurtz]

May the path touch the circle? Must the path be straight line segments? Can part of the path be on the circle?

 

[JDStupi]

Hint 1: circles and lines have equations that define them

Hint 2: a tangent to a circle forms a right-angle with the radius of the circle at that point.

Although in your diagram CDA doesn't look like a right-angle triangle it is!

Forgive me if I'm being dense, but I still can't figure out how to use this. The point D is not given. All I have is the origin, my guess at a point E=(12,y) and my need for a point,D=(x,y), tangent to the circle and on the line that runs from the origin to E.

I tried using your second triangle hint, but knowing only the length of CA and nothing about D, I couldn't put the pieces together.

May the path touch the circle? Must the path be straight line segments? Can part of the path be on the circle?

There is nothing that says the path must be straight line segments, nor that the path can not travel on the arc of the circle, it simply says that it cannot pass through the circle. My first assumption was that that straight line solution would be the shortest because if I traveled along the arc of the circle I would end up covering more distance.

I can see how you can travel from roughly the origin to D and then follow the arc to a point in the first quadrant of the circle and then minimize the distance from that point to the destination (12,16), but that seemed to involve a greater distance as the path must travel a distance upwards while arcing, rather than just moving straight up.

 

[LCKurtz]

If you imagine angling the line BF a bit to the left as you move down so it is tangent to the circle at a point P, isn't it obvious that the path B to P arc to D straight to the origin C would be the shortest? You still have to figure out where C is, where P is, and how long the arc is to add to the straight line segments.

 

[epenguin]

Draw another tangent from C to the circle, tangent at H, say, and meeting GB at I. Then, if we are limited to straight line paths, I think you have a very simple argument from the triangle inequality that CHIGB is a shortest path. There is another path, drawing a new tangent from B, but I think it has the same length.

If instead non-straight line paths are allowed, I think it is not hard to see by the same sort of argument, but harder to justify, that the same tangents
plus an arc HG of the circleIs the shortest path. Something like this should even hold when the arc HG is not only a circle, but convex everywhere.