Question

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. $x-y=0$
• C. $xy=0$
• D. None of these

Answer 1

Answer: (C)

$xy=0$

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

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

therefore, $k=mh+a\sqrt{1+m^2}={(k-mh)}^2=a^2(1+m^2)$ this is a quadratic equation in $m$.

Let the two roots be $m_1$ and $m_2$, then

$m_1+m_2=\frac{2hK}{K^2-a^2}$

But $\tan \alpha =m_1$, $\tan \beta =m_2$ and it is given that

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

$\therefore \frac{1}{m_1}+\frac{1}{m_2}=0$

$\Rightarrow m_1+m_2=0$

$\Rightarrow \frac{2hK}{K^2-a^2}=0$

$\Rightarrow hK=0$

Hence, the locus of $(h,K)$ is $xy=0$.