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 \[\frac{1}{2}\]You need to login to perform this action.
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