Monomath – CBSE | GSEB | IB | JEE Maths by Ajay Yadav (Math King of Katargam) – 15+ Years Experience. Study Notes, Video Lessons, Formula Sheets & PYQs.
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Circles

Previous Year Questions (2022-2024)

📅 Year 2022

  1. PQ and PR are tangents from P to circle with center O. Find ∠QOR if ∠QPR = 60°.
    Show Solution
    OQ ⟂ PQ, OR ⟂ PR (radius ⟂ tangent)\n∠OQP = ∠ORP = 90\nIn quadrilateral OQPR: ∠QOR + 90 + 60 + 90 = 360\n∠QOR = 120
  2. In a circle of radius 5 cm, tangent at point P meets line through O at Q such that OQ = 13 cm. Find PQ.
    Show Solution
    OP ⟂ PQ\nPQ2 = OQ2 - OP2 = 169 - 25 = 144\nPQ = 12 cm

📅 Year 2023

  1. PQ and PR are tangents from P to circle with center O. Find ∠QOR if ∠QPR = 60°.
    Show Solution
    OQ ⟂ PQ, OR ⟂ PR (radius ⟂ tangent)\n∠OQP = ∠ORP = 90\nIn quadrilateral OQPR: ∠QOR + 90 + 60 + 90 = 360\n∠QOR = 120
  2. In a circle of radius 5 cm, tangent at point P meets line through O at Q such that OQ = 13 cm. Find PQ.
    Show Solution
    OP ⟂ PQ\nPQ2 = OQ2 - OP2 = 169 - 25 = 144\nPQ = 12 cm

📅 Year 2024

  1. PQ and PR are tangents from P to circle with center O. Find ∠QOR if ∠QPR = 60°.
    Show Solution
    OQ ⟂ PQ, OR ⟂ PR (radius ⟂ tangent)\n∠OQP = ∠ORP = 90\nIn quadrilateral OQPR: ∠QOR + 90 + 60 + 90 = 360\n∠QOR = 120
  2. In a circle of radius 5 cm, tangent at point P meets line through O at Q such that OQ = 13 cm. Find PQ.
    Show Solution
    OP ⟂ PQ\nPQ2 = OQ2 - OP2 = 169 - 25 = 144\nPQ = 12 cm