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Q.
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 :
Equation of tangent to circle
$x^{2}+y^{2}=16$ is $y=m x \pm 4 \sqrt{m^{2}+1}$
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 $P A$ or $P B$ will be $y=m x \pm h$
They intersect at $x$ -axis, where
$0=m x \pm h \Rightarrow m x=F h$
or $x=F \frac{h}{m} \Rightarrow A B=\frac{2 h}{|m|}$
$\therefore $ Area of $\Delta P A B=\frac{1}{2}\left(\frac{2 h}{|m|}\right) \cdot h=\frac{h^{2}}{|m|}$
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 $\Delta P A B=\frac{4 h^{2}}{\sqrt{h^{2}-16}}=f(h)$ (say)
$\Rightarrow f'(h)=\frac{4\left(h^{3}-32 h\right)}{\left(h^{2}-16\right)^{3 / 2}}=0$
$\Rightarrow h=4 \sqrt{2}$
$\therefore $ For minimum Area, $h=4 \sqrt{2}$
