Maximising the area of a triangle where the vertices lie on circles of certain radius.

Question

Two internally tangent circles $A$ and $B$ have radius $r_1$ and $r_2$ respectively, where $r_2>r_1$. Let $X$ be the internal tangency point, $C$ be a point on $A$ that is not $X$, and $D$ be a point on $B$ that is not $X$. Find the maximum area of $\triangle CDX$.

I've tried connecting $DA$ and $CA$ and then applying the law of sines, but this isn't very helpful - the only properties of the sidelengths I can find are some simple inequalities. I've tried using various projections, and messing with simple cases of $r_1,r_2$ to see if I can generalize. Perhaps a geometric step might be helpful here?

EDIT: I would appreciate if you just provided a hint in the main answer, and provided the rest of the solution in spoilers

Answer 1

I find a simple way: set tangent point is O, set one line $y=ax$, another line is $y=-bx$, circles are $(x-r_1)^2+y^2=r_1^2 ,(x-r_2)^2+y^2=r_2^2$

you get two points at tow circles,take area formula you will get $S=\dfrac{2r_1r_2(a+b)}{(1+a^2)(1+b^2)}$

so the problem become to find max of $ \dfrac{(a+b)}{(1+a^2)(1+b^2)}$ which is easy now.

Answer 2

Found a nice solution!

Extend $XC$ to touch circle $B$ at point $E$.Note that $\frac{|\triangle XCD|}{|\triangle XED| }= \frac{XC}{XE}$ by shared altitude, and $\frac{XC}{XE}=\frac{r_1}{r_2}$ because $\triangle XAC\sim \triangle XBE$. Hence by maximizing the area of $\triangle XED$, we maximise the area of $\triangle XCD$. It's known that the maximum area of a triangle inscribed in a circle is equilateral - and so we find$$\max\{|\triangle XED|\}=\frac{3r_2^2\sqrt{3}}{4}$$ from which it follows $\max\{\triangle XCD\}=\frac{r_1r_2\sqrt{3}}{4}$.