If a 2b 0 then the positive value of m for which y = mx - b√(1+m²) is a common tangent to x² + y

Q: Why do we have an absolute value in the distance formula? Does it really matter?

It matters immensely! The absolute value accounts for the fact that the line could be on either side of the center. In this specific geometry, it leads to two potential cases. Ignoring it would mean missing half the story of the line's orientation. Always respect the absolute value; it is the mathematical guardian of geometric reality.

Q: Why did we reject the case where $m=0$?

Great observation! When we solved the absolute value equation, one branch resulted in $m=0$. Geometrically, $m=0$ represents a horizontal tangent. While valid for some configurations, the problem asks for a specific positive value, and $m=0$ doesn't satisfy the transverse nature of the tangent we are looking for in this specific setup. It is a trivial solution that we must discard to find the 'real' answer.

Q: Is there a faster way to do this without the distance formula?

You could use the condition of tangency for the second circle directly, but the distance formula is the most robust and 'JEE-proof' method. It connects the algebra of the line to the geometry of the circle perfectly. Stick to the fundamentals; they never fail.

Question

If $a > 2b > 0$ then the positive value of $m$ for which $y = mx - b\sqrt{1+m^2}$ is a common tangent to $x^2 + y^2 = b^2$ and $(x-a)^2 + y^2 = b^2$ is

Options:

(A) $\frac{2b}{\sqrt{a^2-4b^2}}$

(B) $\frac{\sqrt{a^2-4b^2}}{2b}$

(C) $\frac{2b}{a-2b}$

(D) $\frac{b}{a-2b}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast coordinate plane. Before you lie two circles, $C_1$ and $C_2$.

The first, $C_1$, is anchored at the origin $(0,0)$ with a radius $b$. The second, $C_2$, is shifted along the x-axis to a center at $(a,0)$, also with radius $b$.

We are given the condition $a > 2b$. This tells us that these two circles are completely separated, with a gap between them. Our mission is to find the slope $m$ of a common tangent line that bridges these two circles.

### The Tangent as a Bridge

We are given the equation of our line: $y = mx - b\sqrt{1+m^2}$. This is the standard form of a tangent to the circle $x^2 + y^2 = b^2$.

It is already 'pre-tuned' to touch the first circle. Our challenge is to ensure it also touches the second circle.

To do this, we must transform our line into the general form: 
$$mx - y - b\sqrt{1+m^2} = 0$$
This is the 'language' the distance formula understands.

### The Heart of the Problem: The Distance Formula

The core principle of coordinate geometry is that a line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is exactly equal to the radius.

For our second circle $C_2$, the center is at $(a, 0)$ and the radius is $b$. Applying the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, we get:

$$\frac{|m(a) - (0) - b\sqrt{1+m^2}|}{\sqrt{m^2 + (-1)^2}} = b$$

This equation is the bridge. It connects the algebraic slope $m$ to the geometric reality of the circle.

### The Algebra of Truth

Multiplying both sides by the denominator $\sqrt{m^2 + 1}$, we arrive at: 
$$|am - b\sqrt{1+m^2}| = b\sqrt{1+m^2}$$

The absolute value splits our path into two distinct possibilities:
1. $am - b\sqrt{1+m^2} = b\sqrt{1+m^2}$
2. $am - b\sqrt{1+m^2} = -b\sqrt{1+m^2}$

Analyzing Case 2, it simplifies to $am = 0$. Since $a > 0$, this forces $m = 0$. Because $m=0$ represents a horizontal line rather than the transverse tangent we seek, we discard it.

### The Final Triumph

We now focus on Case 1: $am = 2b\sqrt{1+m^2}$. To isolate $m$, we square both sides:

$$(am)^2 = (2b\sqrt{1+m^2})^2$$

This yields $a^2m^2 = 4b^2(1 + m^2)$. Expanding the right side gives:

$$a^2m^2 = 4b^2 + 4b^2m^2$$

Bringing all the $m^2$ terms to one side, we have $m^2(a^2 - 4b^2) = 4b^2$. Solving for $m$, we find the final result:

**$$m = \frac{2b}{\sqrt{a^2 - 4b^2}}$$**

This result is elegant, precise, and perfectly derived. You have successfully bridged the two circles.

Animated Solution for Mathematics - Circles: If $a > 2 b > 0$ then the positive value of $m$ for which $y = m x - b 1 + m^{2}$ is a common tangent to $x^{2} + y^{2} = b^{2}$ and $(x - a)^{2} + y^{2} = b^{2}$ is

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Visualized Solution

Visualizing the First Circle $C_{1}$

Visualizing the Second Circle $C_{2}$

Introducing the Tangent Line

Tangency Condition for Circle 2

Setting up the Distance Equation

Simplifying the Equation

Splitting into Cases

Analyzing Case 2 (The Trivial Case)

Analyzing Case 1

Squaring Both Sides

Expanding and Grouping $m^{2}$

Solving for $m$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent as a Bridge

The Heart of the Problem

The Algebra of Truth

The Final Triumph

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a>2b>0 then the positive value of m for which y=mx−b1+m2​ is a common tangent to x2+y2=b2 and (x−a)2+y2=b2 is