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

  $xy=0$
• D

  None of these

Solution

Equation of tangent in slope form



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

${\left(k-mh\right)}^2=a^2 \left(1+m^2\right)$

$m^2 \left(h^2-a^2\right)-2mhk+k^2-a^2=0$

$m_1+m_2= \frac{2hk}{h^2-a^2}, m_1 m_2= \frac{k^2-a^2}{h^2-a^2}$

$\cot \alpha +\cot \beta =0$

$\frac{1}{\tan \alpha }+\frac{1}{\tan \beta }=0 \Rightarrow \frac{\tan \alpha +\tan \beta }{\tan \alpha \tan \beta }=0$

$2hk=0\Rightarrow xy=0$

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