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. \[\therefore \]\[m\ne \pm 2\]You need to login to perform this action.
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