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.

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.
