Animated Solution for Mathematics - Circles: Comprehension Passage

We begin with circle $C_1$ centered at the origin $O(0,0)$ with radius $r_1=2$. The equation of this circle is $x^2+y^2=4$.

For a point $P(\sqrt{3},1)$ on the circle, the equation of the tangent $PT$ is given by the formula $x{x}_1+y{y}_1=a^2$. Substituting the coordinates of $P$, we obtain:

The slope of the tangent $PT$ is $m_1=-\sqrt{3}$. Since line $L$ is perpendicular to $PT$, its slope $m_2$ must satisfy $m_1\cdot m_2=-1$. Thus, $m_2=\frac{1}{\sqrt{3}}$.

The equation of line $L$ can be written in the form $x-\sqrt{3}y+\lambda =0$. We are given that $L$ is tangent to circle $C_2$, which is centered at $(3,0)$ with radius $r_2=1$.

The perpendicular distance from the center $(3,0)$ to the line $L$ must equal the radius $r_2=1$. Using the distance formula:

Simplifying the denominator, we get $\sqrt{1+3}=2$. This leads to the equation:

Solving for $\lambda$, we find two possible values: $\lambda =-1$ or $\lambda =-5$. Therefore, the two possible lines $L$ are $x-\sqrt{3}y-1=0$ and $x-\sqrt{3}y-5=0$.

The distance between the centers $(0,0)$ and $(3,0)$ is $d=3$. Since the sum of the radii is $r_1+r_2=2+1=3$, the circles touch each other externally.

To find the common tangent, we locate the external center of similitude $S$, which divides the line segment joining the centers in the ratio $2:1$ externally:

Any line passing through $S(6,0)$ has the equation $y=m(x-6)$, or $mx-y-6m=0$. Applying the condition that the distance from the origin $(0,0)$ to this line equals $r_1=2$:

Squaring both sides yields $36m^2=4(m^2+1)$, which simplifies to $9m^2=m^2+1$, or $8m^2=1$. Thus, $m=\pm \frac{1}{2\sqrt{2}}$.

Substituting $m$ back into the line equation, we arrive at the final common tangent equation: