Question

Let the tangents drawn to the circle, $x^2+y^2=16$ from the point $P(0,h)$ meet the x-axis at points $A$ and $B$. If the area of $△APB$ is minimum, then $h$ is equal to:

• A. $4\sqrt{2}$
• B. $4\sqrt{3}$
• C. $3\sqrt{2}$
• D. $3\sqrt{3}$

Answer 1

The correct option is A $4\sqrt{2}$


Let $m$ be the slope at $(0,h)$.
$\therefore$ Equation of tangent at $(0,h)$ to the circle is $y=mx+h$
$OM=4$
So, $\frac{h}{\sqrt{1+m^2}}=4$
$\Rightarrow h=4\sqrt{1+m^2}$
$\Rightarrow m=\pm \sqrt{\frac{h^2-16}{16}}$
$\therefore y=\pm \sqrt{\frac{h^2-16}{16}}x+h$
Coordinates of $A$ and $B$ is $\frac{4h}{\sqrt{h^2-16}}$ and $\frac{-4h}{\sqrt{h^2-16}}$ respectively.
$\text{Area}\left(△APB\right)=\frac{1}{2}\times h\times \frac{2\times 4h}{\sqrt{h^2-16}}=S\left(\text{say}\right)$
$\frac{dS}{dh}=0\Rightarrow h=4\sqrt{2}$