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

Solution

Clearly,$\mathrm{ }\mathrm{A}\mathrm{C}\mathrm{P}$ and $\mathrm{B}\mathrm{C}\mathrm{Q}$ are similar triangles.
Hence,
$\frac{AC}{BC}=\frac{r_1}{r_2}$
$\frac{AC}{BC}=2$
$AC=2BC$ ...(i)

Now, since $B$ , is image of $A$ in line $8x-6y=23$
$AD=BD$
$AB=2AD$ ...(ii)
$AD=$ distance of $A$ from line
$8x- 6y=23$
$AD=\left|\frac{16-16-23}{10}\right|=\frac{5}{2}$
Now as $AC=2BC$
$\Rightarrow AC=2(AC-AB)$
$\Rightarrow AC=2AB$
$\Rightarrow AC=4AD$ (from (ii))
$AC=4\times \frac{5}{2}$
$AC=10$

Hence, the correct answer is 10.