Question

Set of values of $m$ for which two points $P$ & $Q$ 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. $\{-\sqrt{3},\sqrt{3}\}$

Answer 1

The correct option is A $(-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )-\{-2,2\}$

Since $\angle APB=\angle AQB=\frac{\pi }{2}$

So, $y=mx+8$ intersects the circle whose diameter is $AB$.
Equation of circle is $x^2+y^2=16$
$CD<4$
$\Rightarrow \frac{8}{\sqrt{1+m^2}}<4$
$\Rightarrow 1+m^2>4$

$\Rightarrow m\in (-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$
When $m=\pm 2,\angle AQB\ne \frac{\pi }{2}$