To solve this problem, we need to find the locus of the point \( P \) where the tangents to the circle \( x^2 + y^2 = a^2 \) intersect. The tangents have inclinations \( \alpha \) and \( \beta \), and it's given that \( \cot \alpha + \cot \beta = 0 \).
The equation of a tangent to the circle \( x^2 + y^2 = a^2 \) at an angle \( \theta \) is given by:
\(y = mx + \sqrt{a^2 (1 + m^2)}\)
where \( m = \tan \theta \). Here, we have two tangents with slopes \( m_1 = \tan \alpha \) and \( m_2 = \tan \beta \).
For the tangents to intersect at \( P(x, y) \), the equations of the tangents can be written as:
\(y = m_1 x + \sqrt{a^2 (1 + m_1^2)}\)
\(y = m_2 x + \sqrt{a^2 (1 + m_2^2)}\)
Setting these equal to find the intersection point:
\(m_1 x + \sqrt{a^2 (1 + m_1^2)} = m_2 x + \sqrt{a^2 (1 + m_2^2)}\)
\((m_1 - m_2) x = \sqrt{a^2 (1 + m_2^2)} - \sqrt{a^2 (1 + m_1^2)}\)
Given \( \cot \alpha + \cot \beta = 0 \), we can deduce that:
\(\cot \alpha = -\cot \beta \Rightarrow \frac{1}{\tan \alpha} = -\frac{1}{\tan \beta} \Rightarrow m_1 m_2 = -1\)
This implies \( \alpha \) and \( \beta \) are complementary angles, so \( m_1 \) and \( m_2 \) are negative reciprocals.
From the condition \( m_1 m_2 = -1 \), the system of equations simplifies, suggesting the lines are perpendicular. Thus, the product of the coordinates of the intersection \( P(x, y) \) must satisfy:
\(m_1 m_2 = \frac{y^2}{x^2} = -1 \Rightarrow xy = 0\)
The locus of the point \( P(x, y) \) where the tangents intersect is therefore given by \(xy = 0\), which means either \( x = 0 \) or \( y = 0 \).