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 $10\sqrt {3}$ , $10\sqrt {5}$ , $10\sqrt {3} +\frac {5\pi} {3}$ ,
$\frac {40*\sqrt{3}}{3}$
On
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:
$$ 2 (x-6) + 2 (y-8) y^{'}=0$$
$$ 3x+4 y= 50 $$
whose intersections with circle obtained when solved are $ (10,5),(2,11) $ Now discard second wrong solution and find minor arc length shown.
The path that you are taking has length $= 10 + 5\pi = … = 25.7050$.
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 $\dfrac {40 \sqrt 3}{3} = … = 23.0940$.