question_answer

Set of values of \[m\] for which two points \[P\] and \[Q\] lie on the line\[y=mx+8\]so that\[\angle APB=\angle AQB=\frac{\pi }{2}\]\[A\equiv (-4,\,\,0),\,\,\,B\equiv (4,\,\,0)\]is-

A) \[(-\infty ,\,\,-\sqrt{3})\cup (\sqrt{3},\,\,\infty )-\{-2,\,\,2\}\]

B) \[[-\sqrt{3},\,\,-\sqrt{3}]-\{-2,\,\,2\}\]

C) \[(-\infty ,\,\,-1)\cup (1,\,\,\infty )-\{-2,\,\,2\}\]

D) \[\left\{ -\sqrt{3},\,\,\sqrt{3} \right\}\]

Correct Answer: A

Solution :

 Since,\[\angle APB=\angle AQB=\frac{\pi }{2}\]so\[y=mx+8\] intersect the circle whose diameter is\[AB\]. Equation of circle is\[{{x}^{2}}+{{y}^{2}}=16\]             \[CD<4\] \[\Rightarrow \]\[\frac{8}{\sqrt{1+{{m}^{2}}}}<4\Rightarrow 1+{{m}^{2}}>4\] \[\Rightarrow \]\[m\in (-\infty ,\,\,-\sqrt{3})\cup (\sqrt{3},\,\,\infty )\] If the line passing throw the point\[A(-4,\,\,0),\,\,B(4,\,\,0)\]then \[\angle APB=\angle AQB=\frac{\pi }{2}\] does not formed. \[\therefore \]    \[m\ne \pm 2\]