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Q.
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 $x^2+y^2+12 x-20 y+120=0$.
Application of Derivatives
Solution:
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{d y}{d a}=\frac{2(a-2)}{2 \sqrt{(a-2)^2+25}}+\frac{2(a+6)}{2 \sqrt{(a+6)^2+100}} $
$\frac{\text { dy }}{d a} \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 $
$=\frac{17}{3}+\frac{\sqrt{1156}}{3}-4=\frac{17}{3}+\frac{34}{3}-4=17-4=13$
