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 $\Delta APB$ is minimum, then $h$ is equal to :

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

Answer 1

Answer: (D)

Equation of tangent to circle
$x^2+y^2=16$ is $y=mx\pm 4\sqrt{m^2+1}$
It passes through $P(0,h);h>0$
$\Rightarrow h=4\sqrt{m^2+1}$
$\Rightarrow 4\sqrt{m^2+1}=h$
$\therefore$ Equation of tangent $PA$ or $PB$ will be $y=mx\pm h$
They intersect at $x$ -axis, where
$0=mx\pm h\Rightarrow mx=Fh$
or $x=F\frac{h}{m}\Rightarrow AB=\frac{2h}{|m|}$
$\therefore$ Area of $\Delta PAB=\frac{1}{2}\left(\frac{2h}{|m|}\right)\cdot h=\frac{h^2}{|m|}$
Also $\frac{|h|}{\sqrt{m^2+1}}=4$
$\Rightarrow \sqrt{m^2+1}=\frac{|h|}{4}$
$\Rightarrow m^2+1=\frac{h^2}{16}$
$\Rightarrow m^2=\frac{h^2-16}{16}$
$\Rightarrow |m|=\frac{\sqrt{h^2-16}}{4}$
$\therefore$ Area of $\Delta PAB=\frac{4h^2}{\sqrt{h^2-16}}=f(h)$ (say)
$\Rightarrow f^{\prime }(h)=\frac{4\left(h^3-32h\right)}{{\left(h^2-16\right)}^{3/2}}=0$
$\Rightarrow h=4\sqrt{2}$
$\therefore$ For minimum Area, $h=4\sqrt{2}$