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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Mathematics
Chapter
Circle

JEE Main 2017 · 8 April · Online · Mathematics · Circle

If a point P(0, - 2) and Q is any point on the circle, x ^ 2 + y ^...

If a point P(0,−2)P\left(0, - 2\right) and QQ is any point on the circle, x2+y2−5x−y+5=0{x}^{2} + {y}^{2} - 5x - y + 5 = 0, then the maximum value of (PQ)2{\left(PQ\right)}^{2} is

Options

  1. A8+538 + 5\sqrt{3}
  2. B47+1062\dfrac{47 + 10\sqrt{6}}{2}
  3. C14+5314 + 5\sqrt{3}
  4. D25+62\dfrac{25 + \sqrt{6}}{2}

Correct answer

C. 14+5314 + 5\sqrt{3}

Step-by-step solution

Given circle x2+y2−5x−y+5=0{x}^{2} + {y}^{2} - 5x - y + 5 = 0
⇒(x−52)2+(y−12)2=32⇒ {\left(x - \dfrac{5}{2}\right)}^{2} + {\left(y - \dfrac{1}{2}\right)}^{2} = \dfrac{3}{2}
On circle Q=(52+32cosθ,12+32sinθ)Q = ( \dfrac{5}{2} + \sqrt{\dfrac{3}{2}}cos\theta, \dfrac{1}{2} + \sqrt{\dfrac{3}{2}}sin\theta )

PQ2=(52+32cosθ)2+(52+32sinθ)2P{Q}^{2} = {\left(\dfrac{5}{2} + \sqrt{\dfrac{3}{2}}cos \theta \right)}^{2} + {\left(\dfrac{5}{2} + \sqrt{\dfrac{3}{2}}sin \theta \right)}^{2}

PQ2=252+32+532(cosθ+sinθ)P{Q}^{2} = \dfrac{25}{2} + \dfrac{3}{2} + 5\sqrt{\dfrac{3}{2}}\left(cos\theta + sin\theta \right)

=14+532(cosθ+sinθ)= 14 + 5\sqrt{\dfrac{3}{2}}\left(cos \theta + sin \theta \right)

∴∴Maximum value of (PQ)2=14+532(2){\left(PQ\right)}^{2} = 14 + 5\sqrt{\dfrac{3}{2}}\left(\sqrt{2}\right)

=14+53= 14 + 5\sqrt{3}.
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