Let the point B be the reflection of the point A(2, 3) with respect to the line 8x - 6y - 23 = 0. Le

Q: Wait, why is the ratio external? I thought tangents usually divide things internally.

That is a fantastic question! The distinction lies in the type of tangent. A *Direct* Common Tangent (DCT) keeps both circles on the same side, which forces the intersection point $C$ to lie *outside* the segment connecting the centers, hence an external division. An *Internal* Common Tangent would pass between the circles, leading to internal division. Always sketch the circles to confirm!

Q: Why is the distance $AB$ equal to $2d$?

Think of the line $8x - 6y - 23 = 0$ as a mirror. By definition, the reflection $B$ of point $A$ is located such that the mirror line is the perpendicular bisector of the segment $AB$. If the perpendicular distance from $A$ to the line is $d$, then the distance from the line to $B$ must also be $d$. Thus, the total distance $AB = d + d = 2d$.

Q: Can we just find the coordinates of $B$ and $C$ to solve this?

You absolutely could! You could find the reflection $B$ using the formula $\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2\frac{ax_1 + by_1 + c}{a^2 + b^2}$, and then find the intersection of the line $AB$ and the tangent $T$. However, in JEE Advanced, time is your most precious resource. Using the geometric property of the DCT is significantly faster and less prone to calculation errors!

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 the line passing through $A$ and $B$, then the length of the line segment $AC$ is

Sigma Percentile · Accepted Answer

### Analyzing the Reflection

Imagine you are standing in a coordinate plane, looking at a point $A(2, 3)$ and a line $L$ defined by $8x - 6y - 23 = 0$. This line acts as a mirror. When we reflect point $A$ across this mirror to find point $B$, we perform a geometric transformation that preserves the distance to the mirror.

To find the perpendicular distance $d$ from point $A$ to line $L$, we employ the standard formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting our specific values into the formula:

$$d = \frac{|8(2) - 6(3) - 23|}{\sqrt{8^2 + (-6)^2}}$$

Simplifying the numerator, we find $|16 - 18 - 23| = |-25| = 25$. The denominator is $\sqrt{64 + 36} = \sqrt{100} = 10$. Thus, the distance $d = 2.5$. Since $B$ is the reflection of $A$, the total distance $AB$ is $2d$, which gives us **$AB = 5$**.

### The Circles and the Tangent

Now, we introduce two circles, $\Gamma_A$ and $\Gamma_B$, centered at $A$ and $B$ with radii $r_A = 2$ and $r_B = 1$, respectively. We seek a Direct Common Tangent $T$ that touches both circles such that they lie on the same side of $T$.

This tangent intersects the line passing through the centers $A$ and $B$ at a point $C$. The Direct Common Tangent possesses a beautiful property: it intersects the line of centers at a point $C$ that divides the segment $AB$ externally in the ratio of the radii.

Mathematically, this is expressed as:

$$\frac{AC}{BC} = \frac{r_A}{r_B}$$

Substituting our radii, we have $\frac{AC}{BC} = \frac{2}{1}$, which implies **$AC = 2BC$**.

### The Final Synthesis

We know that points $A$, $B$, and $C$ are collinear. Because $C$ is the external division point, the segment $AC$ is the sum of the segment $AB$ and the segment $BC$.

This gives us the linear equation:

$$AC = AB + BC$$

We previously calculated $AB = 5$. Substituting $AC = 2BC$ into our collinearity equation, we get:

$$2BC = 5 + BC$$

Solving for $BC$, we find **$BC = 5$**. Finally, the length of the line segment $AC$ is $2BC$, which results in:

**$AC = 10$**

This problem demonstrates how coordinate geometry and circle properties intertwine. By visualizing the reflection and applying the ratio property of the Direct Common Tangent, we bypassed tedious coordinate calculations to arrive at the elegant result of **$10$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Given Elements

Distance of a Point from a Line

Substituting the Coordinates

Evaluating the Distance

Locating Reflection Point $B$

Introducing the Circles

The Direct Common Tangent

Property of Direct Common Tangent

Relation Between $A C$ and $B C$

Collinearity of $A$ , $B$ , and $C$

Solving for $B C$

Final Length of $A C$

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Reflection

The Circles and the Tangent

The Final Synthesis

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