Question

Find the length of the shortest path that begins at the point $(2,5)$, touches the $x$-axis and then ends at a point on the circle $x^2+y^2+12x-20y+120=0$.

Answer 1

Answer: 13

Circle with centre $(-6,10)$ and radius $=\sqrt{36+100-120}=4$
Now let $(a,0)$ be a point on the $x$-axis.
If $y$ is the distance from $A$ to $P$ and $P$ to $M$
$y=\sqrt{(a-2{)}^2+25}+\sqrt{(a+6{)}^2+100}-4$
$\frac{dy}{da}=\frac{2(a-2)}{2\sqrt{(a-2{)}^2+25}}+\frac{2(a+6)}{2\sqrt{(a+6{)}^2+100}}$
$\frac{\text{ dy }}{da}\text{ can be zero only if }a-2>0$
$\text{ and }a+6<0\text{ not possible }$
$\text{ or }a-2<0\text{ and }a+6>0$
$\text{ hence }a\in (-6,2)$

solving $\frac{dy}{da}=0$, gives $a=10$ (rejected) or $a=-\frac{2}{3}$ hence
$y_{\min }=\sqrt{\frac{64}{9}+25}+\sqrt{\frac{256}{9}+100}-4$
$=\frac{17}{3}+\frac{\sqrt{1156}}{3}-4=\frac{17}{3}+\frac{34}{3}-4=17-4=13$