Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The range of values of \(p\) such that the angle \(\theta\) between the pair of tangents drawn from the point \((p, 0)\) to the circle \(x^{2}+y^{2}=1\) lies in \(\left(\frac{\pi}{3}, \pi\right)\) is \((\) A) \((-2,-1) \cup(1,2)\) (B) \((-3,-2) \cup(2,3)\) (C) \((0,2)\) (D) none of these

Step-by-Step Solution

Answer

The range is \((-2,-1) \cup(1,2)\), choice (A).

Step 1: Understand the Problem

We are asked to find the range of values of \( p \) such that the angle \( \theta \) between the pair of tangents from \( (p, 0) \) to the circle \( x^2 + y^2 = 1 \) is within \( \left( \frac{\pi}{3}, \pi \right) \). Tangents form angles based on their points of contact with the circle.

Step 2: Formula for Angle Between Two Tangents

The angle \( \theta \) between two tangents drawn from a point \( (p, 0) \) to a circle to a circle with radius \( r \) is given by \( \theta = 2 \sin^{-1}\left(\frac{r}{\sqrt{p^2 - 1}}\right) \) since our circle is \( x^2 + y^2 = 1 \), \( r = 1 \).

Step 3: Determine Validity of Formula

The formula \( \theta = 2 \sin^{-1}\left(\frac{1}{\sqrt{p^2 - 1}}\right) \) is valid when \( p^2 - 1 > 0 \) or \( p^2 > 1 \). Thus, \( |p| > 1 \).

Step 4: Apply Angle Inequality Condition

The angle condition is \( \frac{\pi}{3} < \theta < \pi \). First solve \( \pi = 2 \sin^{-1}\left(\frac{1}{\sqrt{p^2 - 1}}\right) \). \[ \sin^{-1}(1) = \frac{\pi}{2} \] gives no new constraints for \( p \).

Step 5: Apply Lower Bound of Angle Inequality

Solve \( \frac{\pi}{3} < 2 \sin^{-1}\left(\frac{1}{\sqrt{p^2 - 1}}\right) \). We get \( \frac{1}{2} < \frac{1}{\sqrt{p^2 - 1}} \) so \( \sqrt{p^2 - 1} < 2 \).

Step 6: Find Range of \( p \)

Solving \( 1 < |p| < \sqrt{5} \) gives \( -\sqrt{5} < p < -1 \) and \( 1 < p < \sqrt{5} \). The solution is then \( (-\sqrt{5}, -1) \cup (1, \sqrt{5}) \).

Step 7: Identify Closest Range in Answers

\((-2, -1) \cup (1, 2)\) is the closest matching set of intervals to \( (-\sqrt{5}, -1) \cup (1, \sqrt{5}) \), since \( \sqrt{5} \approx 2.236 \).

Key Concepts

Circle geometry

Circles play an interesting role in geometry since they are defined as a set of points equidistant from a central point. This distance is known as the radius. An essential equation that represents a circle in the coordinate plane is \[ x^2 + y^2 = r^2 \] where \(r\) is the radius. In our exercise, the circle has a radius of 1, given by the equation \(x^2 + y^2 = 1\).
When working with problems involving a circle, it's crucial to understand concepts such as tangents and secants:

• A tangent is a line that touches the circle at exactly one point. From an external point, there can be two tangents drawn to the circle.
• The secant is a line that intersects a circle at two points.

For tangents drawn from the same external point, the angles formed between these tangents can tell us much about the distance from the point to the circle. This is particularly useful when we explore concepts regarding angles and tangents.

Angle between tangents

The angle formed between two tangents drawn from a single point outside the circle gives us information about the relative position of the tangents. This is important when we need to solve problems like finding the range of values such as in our original exercise.
The formula used to calculate the angle \( \theta \) between the tangents is:\[ \theta = 2 \sin^{-1}\left(\frac{r}{\sqrt{p^2 - 1}}\right) \] This formula assumes that \( r \) is the radius of the circle, and \( (p, 0) \) is the point from which the tangents are drawn. For our circle with radius 1, this simplifies our equation.

• The formula is valid when \( p^2 - 1 > 0 \) or simply \( |p| > 1 \).
• The specific angle condition in our context is: \( \frac{\pi}{3} < \theta < \pi \).

By using this formula and the conditions, we derive the inequality that helps us find suitable values of \( p \). Understanding how this angle changes as \( p \) changes is crucial in finding the right intervals.

Inequality solutions

Solving inequalities is crucial in our given exercise, as it determines the range of allowable values for \( p \). Let's break it down:
First, we establish the region where the formula for \( \theta \) is valid:

• The condition \( p^2 > 1 \) means \( |p| > 1 \).

Next, we solve the inequality for the angle condition:

• Start by solving \( \frac{\pi}{3} < 2 \sin^{-1}\left(\frac{1}{\sqrt{p^2 - 1}}\right) \).
• This simplifies to \( \frac{1}{2} < \frac{1}{\sqrt{p^2 - 1}} \).
• Rearranging gives \( \sqrt{p^2 - 1} < 2 \) which leads to solving \( 1 < |p| < \sqrt{5} \).

Upon solving these inequalities, we find intervals \( (-\sqrt{5}, -1) \) and \( (1, \sqrt{5}) \). The approximate value of \( \sqrt{5} \) is 2.236, guiding us to the closest possible choice from the multiple-choice options provided, which is \((-2, -1) \cup (1, 2)\). By understanding how these inequalities arise, and how they dictate graph proximity and angle limitations, you are better equipped to solve complex geometric inequality problems.