Q95 / 200 · Manipal (MET) 2013 · Mathematics

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

  • A
    $x+y=0$
  • B
    $x-y=0$
  • Correct answer
    C
    $x y=0$
  • D
    None of these

Solution

Let the coordinates of $P$ be $(h, k)$. Let the equation of a tangent from $P(h, K)$ to the circle $x^2+y^2=a^2$ be $y=m x+a \sqrt{1+m^2}$ Since, $P(h, K)$ lies on $y=m x+a \sqrt{1+m^2}$ therefore, $k=m h+a \sqrt{1+m^2}=(k-m h)^2=a^2\left(1+m^2\right)$ this is a quadratic equation in $m$. Let the two roots be $m_1$ and $m_2$, then $m_1+m_2=\frac{2 h K}{K^2-a^2}$ But $\tan \alpha=m_1, \tan \beta=m_2$ and it is given that $\cot \alpha+\cot \beta=0$ $\begin{array}{ll}\therefore & \frac{1}{m_1}+\frac{1}{m_2}=0 \\ \Rightarrow & m_1+m_2=0 \\ \Rightarrow & \frac{2 h K}{K^2-a^2}=0 \\ \Rightarrow & h K=0\end{array}$ Hence, the locus of $(h, K)$ is $x y=0$.

Hence, the correct answer is option: 'C'.

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