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The length of the shortest path that begins at $(2,5)$ touches the x-axis and then

end at point on the circle $x^2+y^2+12x-20y+120=0$

$\bf{My\; Try::}$ Equation of circle in standard form :: $(x+6)^2+(y-10)^2=4^2$

So any point on the circle is $x+6=4\cos \theta$ and $y-10 = 4\sin \theta$

So point $P(4\cos \theta-6,4\sin \theta+10)$ on the circle

and let $B(x,0)$ be any point on $x-$ axis and Let $A(2,5)$be a fixed point.

SO we have to minimize $AB+BP = \sqrt{(x-2)^2+5^2}+\sqrt{(x+6-4\cos \theta)^2+(4\sin \theta+10)^2}$

Now how can i solve it after that, Help required, Thanks

Community
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juantheron
  • 53,015

2 Answers2

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Just for a few moments, think about what it would take to extend your path so that after touching a point on the circle, it also touches the center of the circle, $(-6,10)$.

The part of the path from the point on the circle to the center would be a straight segment of length $4$, since that is the shape of the shortest path from any point on that circle to that circle's center. So now you have a path from $(2,5)$ to a point on the $x$ axis to a point on the circle to $(-6,10)$.

Will any of the pieces between those mentioned points be curved?

What kind of angle might there on the path where it passes through the point on the circle? Could you choose a different point on the circle so that the path from the point on the $x$ axis to the circle's center is shorter?

Once you have the shortest path from $(2,5)$ to $(-6,10)$ through the mentioned points, consider just the portion of the path that starts at $(2,5)$ and ends at a point on the circle. Is it possible that there could be a shorter path that starts at $(2,5)$ and ends at a point on the circle?

The function to minimize is much simpler than your first attempt.

David K
  • 98,388
1

Let $C(2, -5)$ be the reflection of point $A$ across the $x$-axis. Notice that $AB = CB$, so we now only need to minimize $CB + BP$. But there are no restrictions on the path from $C$ to $P$ (since any such path must eventually cross the $x$-axis), so the shortest path between the two fixed points is a straight line. So we now only need to minimize $CP$.

Let $D = (-6, 10)$ be the center of the circle. Notice that $PD = 4$, so we now only need to minimize $CD - 4$. But $CD$ is a fixed distance, so we conclude that the optimal path length is: $$ \sqrt{(2 + 6)^2 + (-5 - 10)^2} - 4 = 13 $$

Adriano
  • 41,576