question_answer

Set of values of m for which two points P and Q lie on the line y = mx + 8 so that \[\angle APB=\angle AQB=\frac{\pi }{2}\]where \[A\equiv (-4,0),\] \[B\equiv (4,0)\]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)  \[\{-\sqrt{3},\sqrt{3}\}\]

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. \[\therefore \]\[m\ne \pm 2\]