Question

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

Answer 1

Step 1: Finding the value of $AC$:

Given two circles with centre $A$ and $B$ with radii $2$ and $1$ respectively. The equation of the normal is given as $8x-6y-23=0$.

Also $T$ is a common tangent for both the circles with centre $A$ and $B$ and the tangent is intersecting at the point $C$.

By comparing the two triangles $∆APC$ and $∆BQC$, we have

$\Rightarrow \sin \theta =\frac{2}{AC}=\frac{1}{BC}$

$\Rightarrow \frac{BC}{AC}=\frac{1}{2}$

Here $BC$ can be written as $AC-AB$.

$$\begin{gathered} \Rightarrow \frac{AC-AB}{AC}=\frac{1}{2} \\ \Rightarrow 2\left(AC-AB\right)=AC \\ \Rightarrow 2AC-2AB=AC \\ \Rightarrow AC=2AB \end{gathered}$$

Step 2: Finding the length of the line segment:

Since $R$ is the mid-point of the points $A$ and $B$, so $AB$ can be written as $2AR$.

$\Rightarrow AC=2\left(2AR\right)=4AR$

Using distance formula for the line equation $8x-6y-23=0$ and the point $A\left(2,3\right)$,

$$\begin{gathered} \mathbf{\Rightarrow }\mathit{A}\mathit{R}\mathbf{=}\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{\left(A^2+B^2\right)}}\right| \\ \Rightarrow AR=\left|\frac{2\left(8\right)+3\left(-6\right)+\left(-23\right)}{\sqrt{{\left(8\right)}^2+{\left(-6\right)}^2}}\right| \\ \Rightarrow AR=\left|\frac{16-18-23}{10}\right| \\ \Rightarrow AR=\left|-\frac{25}{10}\right| \\ \Rightarrow AR=\frac{5}{2} \end{gathered}$$

Now substitute $AR$ as $\frac{5}{2}$ in the equation $AC=4AR$.

$$\begin{gathered} \Rightarrow AC=4\left(\frac{5}{2}\right) \\ \Rightarrow AC=2\left(5\right) \\ \Rightarrow AC=10 \end{gathered}$$

Therefore, the length of the line segment $AC$ is $10$.