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Question #40541

Difficulty: 8/10 | 🟢 Free

The Question

Let $C_1$ be the circle $x^2 + y^2 = 25$ and $C_2$ be the circle $x^2 + y^2 = 16$ . Let $S$ be the set of all points $P$ on $C_1$ such that the chord of contact of $P$ with respect to $C_2$ intersects $C_2$ at points $A$ and $B$ . If the coordinates of both $A$ and $B$ must satisfy the strict inequality $|x| + |y| < 4\sqrt{2}$ , then the number of points on $C_1$ that are NOT in $S$ is:

Options

A.
8
B.
0
C.
4
D.
16

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Modulus Inequality Logic|Chord of Contact Logic|Closed vs Open Intervals

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle discrete boundary exclusions within continuous geometric domains (the 'Closed vs Open Interval' trap).

The modulus inequality $|x| + |y| < 4\sqrt{2}$ on the circle $x^2 + y^2 = 16$ is satisfied everywhere except at exactly four discrete points where $|x| + |y| = 4\sqrt{2}$ .

Students often overlook this, assuming the strict inequality doesn't exclude any valid poles, leading to an incorrect answer of 0.

To find the excluded poles, the student must pivot to coordinate geometry, realizing that if the chord of contact passes through a boundary point, the pole must lie on the tangent at that point (pole-polar property), yielding exactly 8 excluded points.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	Modulus Inequality Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free