Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$ 
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.

I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-3)^2}=3PA+PB$$
Where $P$ is a generic point on $x^2+y^2=4$ and $A(1,0),B(0,3)$.
So the problem essentially means which point on the circle $x^2+y^2=4$ minimizes $3PA+PB$.
I am really struggling to find the geometrical notion of this and hence unable to solve.
 A: Comment: you may use this modification:
We rewrite final relation as:
$$A=\sqrt{[3(x-1)=a]^2+(3y=b)^2}+\sqrt{(x=a')^2+(y-3=b')^2}$$
and use this inequality:
$\sqrt{a^2+b^2}+\sqrt{a'^2+b'^2}\geq\sqrt{(a+a')^2+(b+b'^2)}$
we get:
$A\geq\sqrt{16(x^2+y^2)-24(x+y)+18}$
$x^2+y^2=4$
$\Rightarrow$
$A\geq \sqrt{82-24(x+y)}$
If $x=y=\sqrt 2$ then $A\geq\sqrt{84-67.7}\approx 4$
Update:Wolfram says minimum is $2\sqrt {10}=6.32$ at $(x,y)=(\frac25+\frac{3\sqrt6}5=1.87, \frac 65-\frac{\sqrt6}5)=0.71$. If we put this in $\sqrt{82-24(x+y)}$ we get 4.47. mind you x and y must suffice $x^2+y^2=4$, and what Wolfram gives does; $1.87^2+0.71^2=4$.Hence 40 can not be minimum .
A: We can parametrize the constraint condition as $$(x,y) = (2 \cos t, 2 \sin t), \quad t \in [0,2\pi),$$ hence we are interested in the extrema of $$f(t) = 3\sqrt{5-4\cos t} + \sqrt{13-12\sin t}.$$  Differentiating to locate critical points, we find $$\sin t \sqrt{13 - 12 \sin t} = \cos t \sqrt{5 - 4 \cos t},$$ or $$\tan^2 t = \frac{5 \sec t - 4}{13 \sec t - 12 \tan t}.$$  Collecting like terms yields $$(12 \tan^3 t - 4)^2 = (13 \tan^2 t - 5)^2 \sec^2 t,$$ hence with $z = \tan t$,
$$\begin{align}
0 &= (13 \tan^2 t - 5)^2 (1 + \tan^2 t) - (12 \tan^3 t - 4)^2 \\
&= 25 z^6 + 39 z^4 + 96 z^3 - 105 z^2 + 9 \\
&= (5z^2 + 6z - 3)(5z^4 - 6z^3 + 18z^2 - 6z - 3).
\end{align}$$
Consequently, the critical values correspond to$$\tan t = z = \frac{-3 \pm 2 \sqrt{6}}{5}.$$  The roots of the quartic factor are all complex-valued.  It follows that $$(\cos t, \sin t) = \left\{ \left( \frac{3 \sqrt{6} - 2}{10}, \frac{3 \sqrt{6} + 2}{10} \right), \left( - \frac{6 + \sqrt{6}}{10}, \frac{6 - \sqrt{6}}{10} \right) \right\},$$ and $$f(t) = \left\{ \frac{\sqrt{101 + 6 \sqrt{6}} + 9 \sqrt{3} - 3 \sqrt{2}}{\sqrt{5}}, 2 \sqrt{10} \right\}.$$  The first one is the maximum and the second is the minimum.  Attached is a plot of $f$:

A: Nice problem! Here is a solution using geometry.
For the moment, lets forget $P$ is lying on the circle $x^2+y^2=4$. Call the origin $O=(0,0)$. $A=(1,0), B=(0,3)$. So $OA=1, OB=3$ and $\angle BOA = 90^\circ$. Draw the quadrilateral $OAPB$.

Note that we have to minimize
$$3\cdot PA+1\cdot PB=OB\cdot PA + OA\cdot PB$$
We know from Ptolemy's inequality that
$$OB\cdot PA + OA\cdot PB \ge AB \cdot OP$$
with equality only when $O,A,P,B$ lie on a circle.
Thus the desired minimum is attained when $P$ is the intersection of circle with diameter $AB$, call it $S_2$ and $S_1:x^2+y^2=4$.
The value of this minimum is $AB\cdot OP $. Since $AB=\sqrt{10}$ and $OP$ is radius of $S_1$, this minimum value is $2\sqrt{10}$.

Though coordinates of $P$ are not needed, they can be quickly found as intersection of common chord of the two circles and the circle $S_1$. Equation for common chord is $x+3y=4$ hence ordinate of $P$ satisfies $5y^2-12y+6=0$ whose only geometrically valid solution is
$$y_P=\frac{6-\sqrt{6}}{5} \Rightarrow x_P=\frac{2+3\sqrt{6}}{5}$$
which matches other answers and WA.
