Length of Line Segment Between Circle Center and Common Tangent Intersection

To find the length of the line segment $AC$, we proceed step-by-step:

1. Calculate the distance between the centers $A$ and $B$: The point $A$ is given as $(2,3)$ and the line of reflection is $8x - 6y - 23 = 0$. The perpendicular distance $d$ from point $A(2,3)$ to the line $8x - 6y - 23 = 0$ is given by: $d = \frac{|8(2) - 6(3) - 23|}{\sqrt{8^2 + (-6)^2}} = \frac{|16 - 18 - 23|}{\sqrt{64 + 36}} = \frac{|-25|}{10} = 2.5$

   Since $B$ is the reflection of point $A$ across the line, the line segment $AB$ is perpendicular to the line of reflection, and the distance between $A$ and $B$ is twice the distance $d$: $AB = 2d = 2 \times 2.5 = 5$

2. Properties of the common tangent $T$ and point $C$:

   • The radius of circle $\Gamma_A$ centered at $A$ is $r_A = 2$.
   • The radius of circle $\Gamma_B$ centered at $B$ is $r_B = 1$.
   • Since both circles lie on the same side of the common tangent $T$, $T$ is a direct (external) common tangent.

   The intersection $C$ of a direct common tangent with the line joining the centers $A$ and $B$ is the external center of similitude. Thus, point $C$ divides the line segment $AB$ externally in the ratio of the radii: $\frac{AC}{BC} = \frac{r_A}{r_B} = \frac{2}{1} = 2$

3. Calculate the length $AC$: Since $r_A > r_B$, point $C$ lies on the extension of line segment $AB$ beyond point $B$. Therefore, we have: $AC = AB + BC$

   Using $\frac{AC}{BC} = 2$, we get: $AC = 2 \cdot BC$

   Substituting this into $AC = AB + BC$: $2 \cdot BC = AB + BC \implies BC = AB = 5$

   Hence, the length of segment $AC$ is: $AC = 2 \cdot BC = 2 \times 5 = 10$