Circles
A line y = mx + 4 is tangent to the circle x^2 + y^2 = 4; what is the value of the slope m that makes this tangency possible?
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About This Question
- Subject
- mathematics
- Chapter
- coordinate geometry
- Topic
- circles
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
±3
Tangency of a line to a circle requires the perpendicular distance from the centre to the line to equal the radius; this distance condition converts a geometric requirement into one algebraic equation. The circle x^2 + y^2 = 4 has centre (0, 0) and radius 2. Writing the line as mx - y + 4 = 0, the distance from the origin is |4| / sqrt(m^2 + 1). Setting this equal to 2 gives 4 / sqrt(m^2 + 1) = 2, so sqrt(m^2 + 1) = 2, hence m^2 + 1 = 4 and m^2 = 3, giving m = ±sqrt(3). Option ±1 would require sqrt(m^2 + 1) = sqrt(2), inconsistent with distance 2. Option ±2 gives sqrt(5), again not matching. Option ±sqrt(2) yields sqrt(3), too small for radius 2. This applies the canonical JEE Advanced tangent-distance criterion. Plausibility check: with m = sqrt(3), the distance is 4/2 = 2 exactly equal to the radius, so the line grazes the circle at a single point, confirming genuine tangency.
This medium difficulty mathematics question is from the chapter coordinate geometry, covering the topic of circles. It appeared in the 2025 exam.
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