Question

Let the point $B$ be the reflection of the point $A(2,3)$ with respect to the line $8x-6y-23=0.$ Let ${\Gamma }_A$ and ${\Gamma }_B$ be circles of radii $2$ and $1$ with centres $A$ and $B$ respectively. Let $T$ be a common tangent to the circles ${\Gamma }_A$ and ${\Gamma }_B$ such that both the circles are on the same side of $T$. If $C$ is the point of intersection of $T$ and line passing through $A$ and $B,$ then the length of the line segment $AC$ is

Answer 1

Given line: $8x-6y-23=0$
Distance of point $A(2,3)$ from the given line
$AR=\left|\frac{8x_1-6y_1-23}{\sqrt{8^2+6^2}}\right|=\left|\frac{8\times 2-6\times 3-23}{\sqrt{8^2+6^2}}\right|=\frac{25}{10}=\frac{5}{2}$

$\therefore AR=\frac{5}{2}$ and $AB=5$
From similar triangle
$\frac{QB}{PA}=\frac{CB}{CA}\Rightarrow \frac{1}{2}=\frac{BC}{AB+BC}\Rightarrow 5+BC=2BC\Rightarrow BC=5\therefore AC=5+5=10$

Alternate Solution:
$C$ is E.C.S (External Centre of Similitude), which divides $\overline{AB}$ in the ratio $r_1:r_2$ externally.
$\therefore \frac{AC}{BC}=\frac{2}{1}$
$\Rightarrow B$ is the midpoint of $\overline{AC}$

$\therefore AC=4AR$
$=4\times \left|\frac{8x_1-6y_1-23}{\sqrt{8^2+6^2}}\right|=4\times \left|\frac{8\times 2-6\times 3-23}{\sqrt{8^2+6^2}}\right|=4\times \frac{25}{10}=10$