question_answer

AB is a chord of the circle \[{{x}^{2}}+{{y}^{2}}=\frac{25}{2}.\]P is a point such that \[PA=4\]and\[PB=3\]. If \[AB=5,\]then distance of P from origin can be

A) \[\frac{9}{\sqrt{2}}\]                 

B)        \[\frac{3}{\sqrt{2}}\]

C) \[\frac{5}{\sqrt{2}}\]                 

D)        \[\frac{7}{\sqrt{2}}\] or \[\frac{1}{\sqrt{2}}\]

Correct Answer: D

Solution :

[d]        \[OA=OB=\]radius \[=\frac{5}{\sqrt{2}}\] and \[AB=5\] \[\Rightarrow \,\,\,\,\angle AOB=\frac{\pi }{2}\Rightarrow \angle OAB=45{}^\circ \] Again \[PA=4,\] \[PB=3\] and                         \[AB=5\Rightarrow \angle BPA=90{}^\circ \] Let \[\angle PAB=\theta \] Then \[O{{P}^{2}}=O{{A}^{2}}+A{{P}^{2}}-2\times OA\times AP\times \cos (45{}^\circ \pm \theta )\] \[=\frac{25}{2}+16-2\times \frac{5}{\sqrt{2}}\times 4\times \frac{1}{\sqrt{2}}\times \left( \cos \theta \mp \sin \theta  \right)\] \[=\frac{57}{2}-20\left( \frac{4}{5}\mp \frac{3}{5} \right)=\frac{49}{2}\] or \[\frac{1}{2}\]