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What is shortest path between two points

$P (0,0)$ and $Q (12,16)$ such that the path doesn't cross the circle

$(x-6)^2 +(y-8)^2 = 25$ ?

Edit Here is a graph of path of length $10+5\pi$

and this value is one of the options in the problem

The others are $10\sqrt {3}$ , $10\sqrt {5}$ , $10\sqrt {3} +\frac {5\pi} {3}$ ,

$\frac {40*\sqrt{3}}{3}$

enter image description here

user373141
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    Have you tried anything, like drawing a picture, perhaps? – David Quinn Apr 08 '17 at 09:48
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    There are some symmetries such that the points and the center of the circle (0,0), (6,8), (12,16) being co-linear, and the points being on perfect opposite sides of the circle, and the linear distance between start and end being 20 while the radius of the circle is 5. Thus you can simplify the numerics by rotations and scaling.

    I'm going to start with the problem of start and end points (0,0) and (4,0) with the circle at (2,0) with radius 1.

     – Squid Apr 08 '17 at 09:50

2 Answers2

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The path that you are taking has length $= 10 + 5\pi = … = 25.7050$. enter image description here Note that the shaded triangle is equilateral and the circle is inscribed in a rhombus of sides $\dfrac {20 \sqrt 3}{3}$; (which can be found through the "30-60-90" special angled triangle).

If we go along the green paths, the distance travelled is $\dfrac {40 \sqrt 3}{3} = … = 23.0940$.

Mick
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HINT

"Does not cross "means also that it is tangential. Start with a sketch. Assume that the shortest path is a circle tangent to it.

So find the tangent slope $ =\dfrac{16}{12} =\dfrac{4}{3} $

Differentiate circle equation and the above slope in:

$$ 2 (x-6) + 2 (y-8) y^{'}=0$$

$$ 3x+4 y= 50 $$

enter image description here

whose intersections with circle obtained when solved are $ (10,5),(2,11) $ Now discard second wrong solution and find minor arc length shown.

Narasimham
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