Question
Download Solution PDFTangent to the circle at points A and B from any external point P is represented by pair of straight lines x2 - 3y2 - 2x + 1 = 0. If O is the center of the circle then ∠ AOB will be
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
1. Tangent is perpendicular to the radius at the point of contact.
2. Angle between two lines: The angle θ between the lines having slope m1 and m2 is given by
3. Sum of interior angles of a quadrilateral = 360°
Calculation:
Given that, tangent to the circle at points A and B is represented by pair of straight lines PA & PB
x2 - 3y2 - 2x + 1 = 0 -----(1)
From equation (1)
x2 - 3y2 - 2x + 1 = 0
⇒ (x - 1)2 = 3y2
⇒ x - 1 = ± √3y
Therefore, eqation of tangent PA and PB are
x + √3y - 1 = 0 ----(2)
x - √3y - 1 = 0 ----(3)
We know that, slope of line ax + by + c = 0 is -a/b. Therefore,
The slope of line (2) and (3) are -1/√3 and 1/√3
Therefore, angle b/w tangents PA & PB
We know that,
∠ O + ∠ A + ∠ P + ∠ B = 360∘
⇒ ∠ O = 360∘ - 180∘ - 60∘
⇒ ∠ O = 120∘
Hence, option 4 is correct.
Last updated on Jul 14, 2025
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