Finding Common Tangent to Two Given Circles

To find a common tangent to the two circles:

1. Circle $C_1$: $x^2 + y^2 = 4$ has center $O_1(0, 0)$ and radius $r_1 = 2$.
2. Circle $C_2$: $(x - 3)^2 + y^2 = 1$ has center $O_2(3, 0)$ and radius $r_2 = 1$.

The distance between the centers is: $d = O_1O_2 = \sqrt{(3-0)^2 + (0-0)^2} = 3$

Since $r_1 + r_2 = 2 + 1 = 3 = d$, the two circles touch each other externally at the point $(2, 0)$.

Thus, there are three common tangents:

1. One transverse common tangent at the point of contact $(2, 0)$, given by the equation: $x = 2$

2. Two direct common tangents which intersect at the external center of similitude $T_e$, given by: $T_e = \left(\frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2}\right) = \left(\frac{2(3) - 1(0)}{2 - 1}, 0\right) = (6, 0)$

Let the equation of a direct common tangent passing through $(6, 0)$ be: $y - 0 = m(x - 6) \implies mx - y - 6m = 0$

Since this line is tangent to $C_1$, the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $r_1 = 2$: $\frac{|m(0) - 0 - 6m|}{\sqrt{m^2 + (-1)^2}} = 2$ $\frac{|-6m|}{\sqrt{m^2 + 1}} = 2$ $36m^2 = 4(m^2 + 1)$ $32m^2 = 4 \implies m^2 = \frac{1}{8} \implies m = \pm \frac{1}{2\sqrt{2}}$

Substituting $m = -\frac{1}{2\sqrt{2}}$ into the equation of the line: $-\frac{1}{2\sqrt{2}}x - y - 6\left(-\frac{1}{2\sqrt{2}}\right) = 0$ $-x - 2\sqrt{2}y + 6 = 0$ $x + 2\sqrt{2}y = 6$

Substituting $m = \frac{1}{2\sqrt{2}}$ gives: $x - 2\sqrt{2}y = 6$

Among the given options, the line $x + 2\sqrt{2}y = 6$ corresponds to option (D).

Therefore, the correct option is D.