I'm having trouble with an excercise that involves calculating the length of a chord by the way its divided by the circles diameter

Question

I'm failing to understand how to even start this exercise. The exercise goes as follows: theres a chord in a circle. It is divided by 2 perpendicular diameters of the circle. It is divided into 1:3:2 the radius of the circle is r.
Calculate the length of the chord.

I have sketched it multiple times trying to draw different triangles inside it (which I'm pretty sure is the way to solve it), but I just don't get anywhere.
The main Problem I think is sketching the Problem differently, to make is make more sense, but I just can't see that.

Can somebody just give me a little hint on how to go forward?

Heres a sketch of the problem:

Answer 1

Call $\;O,\,M,\,N\;$ to the points "center of the circle, intersection of the vertical diameter with the chord, and intersection of the horizontal diameter with the chord", resp .

By the Chords Theorem , we get:

$$\begin{align*}&\text{By the vertical diameter}:\;x\cdot 5x=(r+OM)(\overbrace{2r-(r+OM)}^{=r-OM})\implies 5x^2=r^2-OM^2\\{}\\ &\text{By the horizontal diameter:}\;(x+3x)\cdot2x=r^2-ON^2\implies8x^2=r^2-ON^2\end{align*}$$

and from here (pass your mouse on the following if you want the solution):

$$OM^2=r^2-5x^2\;,\;\;ON^2=r^2-8x^2$$

and by Pythagoras teorem applied in the straight triangle $\;\Delta OMN\;$ (observe that $\;MN=3x$):

$$2r^2-13x^2=9x^2\implies 22x^2=2r^2\implies x=\frac r{\sqrt{11}}>$$

Answer 2

Denote the points on the chord as $A,B,C,D$ from top left to lower right. Let $O$ be the centre of the circle. Let $P$ be the intersection of $OB$ and the parallel to $OC$ through $A$. Let $Q$ be the intersection of $OC$ and the parallel to $OB$ through $D$. Then the three triangles $ABP$, $CBO$, $CDQ$ are similar right triangles. Let $u=AP$, $v=BP$. Then Pythagoras applied to the right triangles $ABP$, $AOP$, and $ODQ$ gives us $$\begin{align}\tag1u^2+v^2&=x^2\\ \tag2u^2+(v+3v)^2&=r^2\\ \tag3(3u+2u)^2+(2v)^2&=r^2 \end{align}$$ We want to find the lenth $\ell=6x$ from this. To do so, combine $(2)$ and $(3)$ suitably to obtain the a multiple of $u^2+v^2$ on the left.

As it turns out, $7(2)+5(3)$ gives us $(7+5)r^2=(7+5\cdot 25)u^2+(7\cdot 16+5\cdot 4)v^2=132\cdot(u^2+x^2)=132 x^2$ and so $r^2=11x^2$.

Finally,

$$\ell=6x=\frac{6r}{\sqrt {11}}.$$