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Tangents are drawn from  P(6,8) to the circle \mathrm{x^2+y^2=r^2} . The radius of the circle such that the area of the \Delta formed by tangents and chord of contact is maximum, is __________.

Option: 1

5


Option: 2

4


Option: 3

3


Option: 4

2


Answers (1)

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\mathrm{\begin{aligned} & \ \tan \theta=\frac{r}{P S}=\frac{r}{\sqrt{100-r^2}}, \\ & \sin \theta=\frac{r}{10}, \quad \cos \theta=\frac{\sqrt{100-r^2}}{10} \end{aligned}}

Now, Area of \triangle P S M=A=\frac{1}{2} S M \times P N

\mathrm{ \begin{aligned} & =\frac{1}{2} \cdot 2 \cdot S N \times P N=S N \times P N=S P \sin \theta \times S P \cos \theta \\\\ & =\left(\sqrt{100-r^2}\right)^2 \times \sin \theta \cos \theta=\left(100-r^2\right) \frac{r}{10} \frac{\sqrt{100-r^2}}{10} \\\\ & \quad=\frac{r\left(100-r^2\right)^{3 / 2}}{100} \\\\ & \therefore \frac{d A}{d r}=\frac{1}{100} \cdot \frac{3}{2}\left(100-r^2\right)^{1 / 2}(-2 r) r+\frac{\left(100-r^2\right)^{3 / 2}}{100}=0 \end{aligned} }

\mathrm{ \Rightarrow\left(100-r^2\right)^{1 / 2}\left(-3 r^2+100-r^2\right)=0 }
Now, r \neq 10 as P is outside the circle

\mathrm{ \begin{aligned} & \Rightarrow 4 r^2=100 \\ & \Rightarrow r^2=25 \Rightarrow r=5 \end{aligned} }

Thus for r=5, A have maximum area since \mathrm{\frac{d^2 A}{d r^2}<0 \, \, for\, \, r=5.}

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manish

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