Problem 20 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a>2 b>0\) then the positive value of \(m\) for which \(y=m x-b \sqrt{1+m^{2}}\) is a common tangent to \(x^{2}+y^{2}=\) \(b^{2}\) and \((x-a)^{2}+y^{2}=b^{2}\) is (A) \(\frac{2 b}{\sqrt{a^{2}-4 b^{2}}}\) (B) \(\frac{\sqrt{a^{2}-4 b^{2}}}{2 b}\) (C) \(\frac{2 b}{a-2 b}\) (D) \(\frac{b}{a-2 b}\)

Step-by-Step Solution

Answer

The positive value of \(m\) is \(\frac{2b}{\sqrt{a^2 - 4b^2}}\).

Step 1: Understand the Problem

The task is to find the positive value of the slope \(m\) for the line \(y = mx - b\sqrt{1 + m^2}\) to be tangent to both circles, \(x^2 + y^2 = b^2\) and \((x-a)^2 + y^2 = b^2\). This means the line will touch the circles at exactly one point, and we need to ensure this condition is met for both circles.

Step 2: Equation of Tangent to First Circle

The given circle is \(x^2 + y^2 = b^2\). The condition for the line \(y = mx - b\sqrt{1 + m^2}\) to be tangent to this circle is: the perpendicular distance from the center \((0,0)\) to the line equals the radius \(b\). This gives us the equation: \[ \frac{|b\sqrt{1 + m^2}|}{\sqrt{1+m^2}} = b \]. Solving, we confirm that the equation holds for any \(m\).

Step 3: Equation of Tangent to Second Circle

For the circle \((x-a)^2 + y^2 = b^2\), the center is \((a, 0)\). The condition for tangency is that the perpendicular distance from \((a, 0)\) to the line \(y = mx - b\sqrt{1 + m^2}\) also equals \(b\). This gives: \[ \frac{|ma - b\sqrt{1 + m^2}|}{\sqrt{1 + m^2}} = b \].

Step 4: Solve for m in Tangency Condition

Solve \( \frac{|ma - b\sqrt{1 + m^2}|}{\sqrt{1 + m^2}} = b \). Rearranging gives: \( |ma - b\sqrt{1 + m^2}| = b\sqrt{1 + m^2} \). This means \( ma = 2b\sqrt{1 + m^2} \) (choosing the positive sign as \(m\) should be positive by the problem statement).

Step 5: Substitute and Simplify

Express \(ma\) as \(ma = b\sqrt{1 + m^2} + b\sqrt{1 + m^2} \). Then substitute \(b\sqrt{1 + m^2} = \frac{ma}{2}\), and solve for \(m\). This gives \[ m = \frac{2b}{a - b}\sqrt{1+m^2} \].

Step 6: Further Simplification

Square both sides to eliminate the square root: \( m^2(a - 2b)^2 = 4b^2 + 4b^2m^2 \). Rearrange to get \( m^2a^2 - 4m^2ab + 4m^2b^2 = 4b^2 + 4b^2m^2 \). Simplify to find \( m = \frac{2b}{\sqrt{a^2 - 4b^2}} \).

Step 7: Ensure Positive m

The positive value for \(m\) is \(m = \frac{2b}{\sqrt{a^2 - 4b^2}}\), which is option \((A)\).

Key Concepts

Equation of Tangent

Understanding the equation of a tangent line to a circle is essential in geometry. When a line is tangent to a circle, it means that the line touches the circle exactly at one point. This is distinct from a secant line, which crosses the circle at two points. To define a tangent line in mathematical terms, we consider the slope of the line and its relationship with the circle's radius.

For a line represented as \(y = mx + c\), to be tangent to a circle with a center at the origin \((0, 0)\) and radius \(r\), the distance from the center of the circle to the line must be equal to the radius \(r\). This means solving the following equation:

• \(\frac{|c|}{\sqrt{1 + m^2}} = r\)

Here, \(m\) is the slope of the tangent, and \(c\) is the y-intercept. This formula is crucial as it provides the condition under which a line is tangent to the circle, ensuring the geometric property is maintained without intersections other than the point of tangency.

Circle Geometry

Circle geometry involves the study of circles and the relationships between different parts of a circle: the radius, diameter, tangent, and center. It is a foundational concept in geometry and has applications in various mathematical problems.

• The equation of a circle in standard form is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
• A tangent to a circle is a line that touches the circle without crossing it, and it forms a right angle with the radius at the point of contact.
• In the exercise, two circles are considered: one with center at \((0,0)\) and the other at \((a,0)\).

When a line is tangent to two different circles, it has to meet the tangency condition simultaneously for both. This requires analyzing the geometric positioning and calculating the correct slope for the tangent that fulfills the conditions of tangency for each circle.

Perpendicular Distance

Perpendicular distance plays a crucial role when analyzing tangents to a circle. It refers to the shortest distance between a point and a line. In the context of geometry, this measure helps determine whether a line indeed becomes a tangent to a circle.

Consider a line \(y = mx + c\). The perpendicular distance \(d\) from a point \((x_1, y_1)\) to this line is given by the formula:

• \(d = \frac{|mx_1 - y_1 + c|}{\sqrt{1 + m^2}}\)

In the exercise, to ensure a line is tangent to the circle, we equate this perpendicular distance to the radius of the circle. This relationship helps solve for the correct slope \(m\) of the tangent line that tangentially touches both given circles in the exercise. By meeting the conditions for both circles, we understand that the line is positioned in such a way that it lightly 'kisses' each circle just once, which is a testament to its tangency.