Verified Solution • 2025–26 Session moderate

QJEEM17Ma5185: JEE Main 2017 Mathematics - Class 11 JEEM17Ma5185 Solution: If a point P has co-ordinates (0, 2) and Q is any point on the…

If a point P has co-ordinates (0, −- 2) and Q is any point on the circle, x2 + y2 −- 5x −- y + 5 = 0, then the maximum value of (PQ)2 is :

Multiple Choice Options

  • A 25+62{{25 + \sqrt 6 } \over 2}
  • B 14 + 535\sqrt 3
  • C 47+1062{{47 + 10\sqrt 6 } \over 2}
  • D 8 + 5 3\sqrt 3

Step-by-Step Verified Solution

Given that x2 + y2 −- 5x −- y + 5 = 0

⇒\Rightarrow    (x −- 5/2)2 −- 254{{25} \over 4} + (y −- 1/2)2 −- 1/4 = 0

⇒\Rightarrow    (x −- 5/2)2 + (y −- 1/2)2 = 3/2

on circle [ Q ≡\equiv (5/2 + 3/2\sqrt {3/2} cos Q, 12{1 \over 2} + 3/2\sqrt {3/2} sin Q)]

⇒\Rightarrow    PQ2 = (52+3/2cos⁡Q)2{\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2} + (52+3/2sin⁡Q)2{\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}

⇒\Rightarrow    PQ2 = 252+32+53/2{{25} \over 2} + {3 \over 2} + 5\sqrt {3/2} (cos Q + sinQ)

= 14 + 5 3/2\sqrt {3/2} (cosQ + sinQ)

∴\therefore    Maximum value of PQ2

= 14 + 5 3/2\sqrt {3/2} ×\times   2\sqrt 2 = 14 + 5 3\sqrt 3

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