如图所示，两个圆 $C_1$ 和 $C_2$ 在 $xy$ 平面内相互相切。$C_1$ 的半径为 4 个单位且与 $x$ 轴相切于点 $D$，$C_2$ 的半径为 3 个单位且与 $y$ 轴相切于点 $E$。连接 $C_2$ 和 $C_1$ 圆心的直线 $AB$ 与 $x$ 轴相交于点 $F$，使得 $\angle BFO = \theta^\circ$。

1. Determine the distance between the centres of the two circles.

Since the circles touch each other externally, the distance between their centres $A$ and $B$ is the sum of their radii:
$AB = 3 + 4 = 7$

2. Express the coordinates of $A$ and $B$ in terms of $OD$ and $OE$.

Circle $C_2$ touches the $y$-axis at $E$, so the $x$-coordinate of $A$ is 3. Thus, $A = (3, OE)$.
Circle $C_1$ touches the $x$-axis at $D$, so the $y$-coordinate of $B$ is 4. Thus, $B = (OD, 4)$.

3. Use trigonometry to find expressions for $OD$ and $OE$.

The line $AB$ makes an angle $\theta$ with the horizontal. By drawing a right-angled triangle with hypotenuse $AB$:
Horizontal distance between $A$ and $B$ is $AB \cos\theta = 7\cos\theta$.
Therefore, $OD - 3 = 7\cos\theta \implies OD = 3 + 7\cos\theta$.

Vertical distance between $A$ and $B$ is $AB \sin\theta = 7\sin\theta$.
Therefore, $OE - 4 = 7\sin\theta \implies OE = 4 + 7\sin\theta$.

4. Show the expression for $ED^2$ using Pythagoras' theorem.

In $\triangle ODE$, $\angle DOE = 90^\circ$.
$ED^2 = OD^2 + OE^2$
$ED^2 = (3 + 7\cos\theta)^2 + (4 + 7\sin\theta)^2$
$ED^2 = (9 + 42\cos\theta + 49\cos^2\theta) + (16 + 56\sin\theta + 49\sin^2\theta)$
$ED^2 = 25 + 49(\cos^2\theta + \sin^2\theta) + 42\cos\theta + 56\sin\theta$
Since $\cos^2\theta + \sin^2\theta = 1$:
$ED^2 = 25 + 49(1) + 56\sin\theta + 42\cos\theta$
$ED^2 = 74 + 56\sin\theta + 42\cos\theta$ (shown)

1. Set up the R-formula equivalence for the trigonometric part of the expression

We need to express $42\cos\theta + 56\sin\theta$ in the form $R\cos(\theta - \alpha)$.
Expanding $R\cos(\theta - \alpha)$ gives $R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 42$
$R\sin\alpha = 56$

2. Calculate the value of R

$R = \sqrt{42^2 + 56^2}$
$R = \sqrt{1764 + 3136} = \sqrt{4900} = 70$

3. Calculate the value of alpha and write the final expression

$\tan\alpha = \frac{56}{42} = \frac{4}{3}$
$\alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.130^\circ \approx 53.1^\circ$ (to 1 d.p.)

Therefore, $ED^2 = 74 + 70\cos(\theta - 53.1^\circ)$.

1. Find the values of $R$ and $\alpha$ from part (ii)

From part (ii), $56\sin\theta + 42\cos\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 42$
$R\sin\alpha = 56$
$R = \sqrt{42^2 + 56^2} = \sqrt{4900} = 70$
$\tan\alpha = \frac{56}{42} = \frac{4}{3} \implies \alpha = 53.130^\circ$
So, $ED^2 = 74 + 70\cos(\theta - 53.130^\circ)$.

2. Find the maximum value of $ED$

For maximum $ED$, $ED^2$ must be maximum.
The maximum value of $\cos(\theta - 53.130^\circ)$ is $1$.
Max $ED^2 = 74 + 70(1) = 144$
Max $ED = \sqrt{144} = 12$

3. Find the value of $\theta$ at which the maximum occurs

The maximum occurs when $\cos(\theta - 53.130^\circ) = 1$.
$\theta - 53.130^\circ = 0^\circ$
$\theta = 53.1^\circ$ (to 1 d.p.)