<s>[INST] length of the shortest path connecting a point and another point on circle. 
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
 [/INST] 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:
$$
\sqrt{(2 + 6)^2 + (-5 - 10)^2} - 4 = 13
$$
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