Circles
From the external point (5, 1) a tangent is drawn to the circle x^2 + y^2 = 9; what is the length of this tangent segment from the point to the circle?
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About This Question
- Subject
- mathematics
- Chapter
- coordinate geometry
- Topic
- circles
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
17
The length of a tangent from an external point (x_1, y_1) to a circle S = 0 equals sqrt(S_1), where S_1 is the circle expression evaluated at that point; this elegant result follows from the Pythagorean relation between radius, tangent, and centre distance. For x^2 + y^2 - 9 = 0, evaluating at (5, 1) gives S_1 = 25 + 1 - 9 = 17, so the tangent length is sqrt(17). Option sqrt(26) is the squared distance from the origin to the point, forgetting to subtract the radius-squared term. Option 4 = sqrt(16) would require S_1 = 16, an arithmetic slip. Option sqrt(35) over-counts the constant. This uses the standard JEE Advanced tangent-length formula. Plausibility check: the point's distance from the centre is sqrt(26) ≈ 5.1, which exceeds the radius 3, confirming the point is external so a real tangent exists, and sqrt(26 - 9) = sqrt(17) is consistent with the right-triangle leg relationship.
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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