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The tangents to \mathrm{x^{2}+y^{2}=a^{2}} having inclinations \alpha and \beta intersect at \mathrm{P}. If \cot \alpha+\cot \beta=0, then the locus of  \mathrm{P} is

Option: 1

\mathrm{x+y=0}


Option: 2

\mathrm{x}-\mathrm{y}=0


Option: 3

\mathrm{x y=0}


Option: 4

None of these


Answers (1)

best_answer

Let the coordinates of \mathrm{P} be (\mathrm{h}, \mathrm{k}), Let the equation of a tangent from \mathrm{P}(\mathrm{h}, \mathrm{k}) to the circle \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}\: be

\mathrm{y}=\mathrm{mx}+\mathrm{a} \sqrt{1+m^{2}}

\mathrm{Since, \mathrm{P}(\mathrm{h}, \mathrm{k})\: lies \: on\: \mathrm{y}=\mathrm{mx}+a \sqrt{1+m^{2}}, } therefore,

\begin{aligned} & \mathrm{k}=\mathrm{mh}+a \sqrt{1+m^{2}} \Rightarrow(k-m h)^{2}=a \sqrt{\left(1+m^{2}\right)} \\ & \Rightarrow \mathrm{m}^{2}\left(\mathrm{k}^{2}-\mathrm{a}^{2}\right)-2 \mathrm{mkh}+\mathrm{k}^{2}-\mathrm{a}^{2}=0 \end{aligned}

This is a quadratic in \mathrm{m}. Let the two roots be \mathrm{m}_{1} and \mathrm{m}_{2}, then
\mathrm{m_{1}+m_{2}=\frac{2 h k}{k^{2}-a^{2}}}

 But\: \tan \alpha=\mathrm{m} 1, \tan \beta=\mathrm{m} 2   and it is given that
\cot \alpha+\cot \beta=0
\therefore \frac{1}{m_{1}}+\frac{1}{m_{2}}=0 \Rightarrow \mathrm{m}_{1}+\mathrm{m}_{2}=0 \Rightarrow \frac{2 h k}{k^{2}-a^{2}}=0
\Rightarrow \mathrm{hk}=0.
Hence, the locus of
(\mathrm{h}, \mathrm{k}) is \mathrm{xy}=0.

Posted by

Anam Khan

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