Chord Distances

Say the perpendicular distance from the centre of the smaller circle to the chord is $a$ and the perpendicular distance from the centre of the larger circle to the chord is $b$ . Let the chord have length $2h$ .

Then $a^2+h^2=5^2,\;\;\;b^2+h^2=7^2,\;\;\;a+b=8$

So $b^2-a^2=24$ ; dividing by $a+b=8$ we get $b-a=3$ .

From these, $a=\frac52$ so $h^2=25-\frac{25}{4}=\frac{75}{4}$ and the length of the chord is $\sqrt{75}=5\sqrt3$ .

We can solve the problem using intersecting chords theorem . Let the length of the common chord be $2c$ and the greatest width of the two intersecting arcs be $a+b$ be divided by the common chord into lengths $a$ and $b$ as shown in the figure. Then we have $5-(a+b) + 7 = 8 \implies a+b = 4 \implies b = 4 -a$ .

By intersecting chords theorem we have:

$\begin{cases} a(14-a) = c^2 \\ b(10-b) = c^2 \end{cases}$

$\begin{aligned} \implies a(14-a) & = b(10-b) \\ 14a - a^2 & = (4-a)(6+a) = 24 - 2a - a^2 \\ 16a & = 24 \\ \implies a & = \frac 32 \end{aligned}$

From $a(14-a) = c^2$ ,

$\begin{aligned} \implies c^2 & = \frac 32 \left(14 - \frac 32 \right) \\ & = \frac {3 \times 25}4 \\ \implies c & = \frac {5\sqrt 3}2 \\ 2c & = \boxed{5\sqrt 3} \end{aligned}$

In $\Delta C_1AC_2$ , applying cosine rule

$\cos \angle C_1AC_2 = \frac{AC_1^2 + AC_2^2 - C_1C_2^2}{2.AC_1.AC_2}$

$\cos \angle C_1AC_2 = \frac{(5)^2 + (7)^2 - (8)^2}{2.(5).(7)}$

$\boxed{\cos \angle C_1AC_2 = \frac17}$

So, $\sin \angle C_1AC_2 = \frac{\sqrt{48}}{7}$

Now, $ar(\Delta C_1AC_2) = ar(\Delta C_1AC_2)$

$\frac12 AC_1.AC_2.sin\angle C_1AC_2 = \frac12 C_1C_2.h$

$\frac12 (5).(7).\frac{\sqrt{48}}{7} = \frac12 (8).h$

$\boxed{h = \frac{5\sqrt3}{2}}$

So,

$\color{#3D99F6}{\boxed{\text{Length of chord } = 2h = 5\sqrt3}}$