JEE Main On 8 April 2017 Question 22
Question: If a point P has co-ordinates (0,-2) and Q is any point on the circle, $ x^{2}+y^{2}-5x-y+5=0, $ then the maximum value of $ {{(PQ)}^{2}} $ is: [JEE Online 08-04-2017]
Options:
A) $ 8+5\sqrt{3} $
B) $ \frac{47+10\sqrt{6}}{2} $
C) $ 14+5\sqrt{3} $
D) $ \frac{25+\sqrt{6}}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{(x-5/2)}^{2}}\frac{-25}{4}+{{(y-1/2)}^{2}}-1/4+5=0 $
$ ={{(x-5/2)}^{2}}+{{(y-1/2)}^{2}}=3/2 $
on circle $ Q=5/2+\sqrt{3/2}\cos Q,\frac{1}{2}+\sqrt{3/2}\sin Q $
$ PQ^{2}={{( \frac{5}{2}+\sqrt{3/2}\cos Q )}^{2}}+{{( \frac{5}{2}+\sqrt{3/2}\sin Q )}^{2}} $
$ PQ^{2}=\frac{25}{2}+\frac{3}{2}+5\sqrt{3/2}(\cos Q+\tan Q) $
$ =14+5\sqrt{3/2}(cos,Q+\tan Q) $
$ =14+5\sqrt{3/2}(\sqrt{2}) $
$ =14+5\sqrt{3} $
