Application Of Derivatives Ques 96
96. For the circle $x^{2}+y^{2}=r^{2}$, find the value of $r$ for which the area enclosed by the tangents drawn from the point $P(6,8)$ to the circle and the chord of contact is maximum.
(2003, 2M)
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Answer:
Correct Answer: 96.5 Units
Solution:
Formula:
Maxima and Minima of functions of one variable :
- To maximise area of $\triangle A P B$, we know that, $O P=10$ and $\sin \theta=r / 10$, where $\theta \in(0, \pi / 2) ……(i)$
$\therefore \quad$ Area $=\frac{1}{2}(2 A Q)(P Q)$
$ \begin{aligned} & =A Q \cdot P Q=(r \cos \theta)(10-O) Q \\ & =(r \cos \theta)(10-r \sin \theta) \\ & =10 \sin \theta \cos \theta\left(10-10 \sin ^{2} \theta\right) \quad \text { [from Eq. (i)] } \end{aligned} $
$\Rightarrow \quad A=100 \cos ^{3} \theta \sin \theta$
$\Rightarrow \quad \frac{d A}{d \theta}=100 \cos ^{4} \theta-300 \cos ^{2} \theta \cdot \sin ^{2} \theta$
Put $\frac{d A}{d \theta}=0$
$\Rightarrow \cos ^{2} \theta=3 \sin ^{2} \theta$
$\Rightarrow \quad \tan \theta=1 / \sqrt{3}$
$\Rightarrow \theta=\pi / 6$
At which $\frac{d A}{d \theta}<0$, thus when $\theta=\pi / 6$, area is minimum
From Eq. (i), $r=10 \sin \frac{\pi}{6}=5$ units
BETA
