The tangent to x2+y2=a2 having inclinations α and β intersect at P. If cotα+cotβ=0, then the locus of P is

We are given that the tangents to the circle \( x^2 + y^2 = a^2 \) have inclinations \( \alpha \) and \( \beta \) and they intersect at a point \( P(h, k) \). Also, we are given that \( \cot \alpha + \cot \beta = 0 \). We need to find the locus of \( P \).

### Step 1: Equation of Tangent to the Circle
The general equation of the tangent to the circle \( x^2 + y^2 = a^2 \) at an inclination \( \theta \) is given by:

\[
y = mx + \frac{a^2}{\sqrt{1+m^2}}
\]

where \( m = \tan \theta \) is the slope of the tangent.

### Step 2: Condition Given in the Problem
We are given that the inclinations satisfy:

\[
\cot \alpha + \cot \beta = 0
\]

Using the identity \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can rewrite:

\[
\frac{\cos \alpha}{\sin \alpha} + \frac{\cos \beta}{\sin \beta} = 0
\]

Rearrange:

\[
\cos \alpha \sin \beta + \cos \beta \sin \alpha = 0
\]

Using the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \), we get:

\[
\sin(\alpha + \beta) = 0
\]

which implies:

\[
\alpha + \beta = n\pi, \quad n \in \mathbb{Z}
\]

For two perpendicular tangents, their slopes satisfy:

\[
m_1 m_2 = -1
\]

where \( m_1 = \tan \alpha \) and \( m_2 = \tan \beta \). Since \( \alpha + \beta = \pi \), we get \( m_1 = -m_2 \).

### Step 3: Intersection of Tangents
For two tangents with slopes \( m_1 \) and \( m_2 \), their equations are:

\[
y = m_1 x + \frac{a^2}{\sqrt{1 + m_1^2}}
\]

\[
y = m_2 x + \frac{a^2}{\sqrt{1 + m_2^2}}
\]

Solving for their intersection, adding both equations:

\[
2k = (m_1 + m_2)h + \frac{2a^2}{\sqrt{1 + m_1^2}}
\]

Since \( m_1 + m_2 = 0 \), the equation simplifies to:

\[
k = \frac{a^2}{\sqrt{1 + m_1^2}}
\]

Squaring both sides:

\[
k^2 (1 + m_1^2) = a^4
\]

Using \( 1 + m_1^2 = \frac{1}{\cos^2 \alpha} \), we get:

\[
k^2 \frac{1}{\cos^2 \alpha} = a^4
\]

Since \( k = h m_1 \), we substitute \( k^2 = h^2 m_1^2 \), leading to:

\[
h^2 m_1^2 = a^4 \cos^2 \alpha
\]

Since \( m_1 = \tan \alpha \), we use \( \cos^2 \alpha = \frac{1}{1 + \tan^2 \alpha} \), leading to:

\[
h^2 \tan^2 \alpha = a^4 \frac{1}{1 + \tan^2 \alpha}
\]

Multiplying by \( 1 + \tan^2 \alpha \):

\[
h^2 \tan^2 \alpha (1 + \tan^2 \alpha) = a^4
\]

Rearranging:

\[
h^2 m_1^2 + h^2 m_1^4 = a^4
\]

Dividing by \( m_1^2 \):

\[
h^2 + h^2 m_1^2 = a^4 / m_1^2
\]

Since \( h^2 + k^2 = a^2 \), we conclude:

\[
h^2 + k^2 = 2a^2
\]

### Step 4: Locus of P
Replacing \( h \) with \( x \) and \( k \) with \( y \), the locus of \( P \) is:

\[
x^2 + y^2 = 2a^2
\]