If `theta_(1), theta_(2)` be the inclination of tangents with x-axis drawn from the point P to the circle `x^(2)+y^(2)=a^(2)`, then the locus of P, if given that `cot theta_(1)+cot theta_(2)=c` is

Options

A. `c(x^(2)-a^(2))=2xy`
B. `c(x^(2)-a^(2))=y^(2)-a^(2)`
C. `c(y^(2)-a^(2))=2xy`
D. none

Answer

To find the locus of the point P from which tangents are drawn to the circle \(x^2 + y^2 = a^2\) such that \( \cot \theta_1 + \cot \theta_2 = c \), we can follow these steps:

### Step 1: Express the condition in terms of tangent
We start with the given condition:
\[ \cot \theta_1 + \cot \theta_2 = c \]
This can be rewritten in terms of tangent:
\[ \frac{1}{\tan \theta_1} + \frac{1}{\tan \theta_2} = c \]
Multiplying through by \(\tan \theta_1 \tan \theta_2\) gives:
\[ \tan \theta_2 + \tan \theta_1 = c \tan \theta_1 \tan \theta_2 \]

### Step 2: Use the tangent properties
Let \(m_1 = \tan \theta_1\) and \(m_2 = \tan \theta_2\). We can express the sum and product of the tangents:
\[ m_1 + m_2 = c m_1 m_2 \]
This implies:
\[ m_1 + m_2 = c \cdot m_1 m_2 \]

### Step 3: Equation of tangents from point P
The equation of the tangents from point \(P(x_1, y_1)\) to the circle \(x^2 + y^2 = a^2\) can be expressed as:
\[ y - y_1 = m(x - x_1) \]
Squaring both sides and substituting into the circle's equation gives:
\[ (y_1 - mx_1)^2 = a^2(1 + m^2) \]

### Step 4: Rearranging the tangent equation
Rearranging gives:
\[ y_1^2 - 2y_1mx_1 + m^2x_1^2 = a^2 + a^2m^2 \]
This can be rewritten as:
\[ (1 + a^2)m^2 - 2y_1mx_1 + (y_1^2 - a^2) = 0 \]
This is a quadratic equation in \(m\).

### Step 5: Finding the roots
Let \(m_1\) and \(m_2\) be the roots of this quadratic equation. By Vieta's formulas:
\[ m_1 + m_2 = \frac{2y_1x_1}{1 + a^2} \]
\[ m_1 m_2 = \frac{y_1^2 - a^2}{1 + a^2} \]

### Step 6: Substituting into the tangent condition
Substituting these into the earlier equation gives:
\[ \frac{2y_1x_1}{1 + a^2} = c \cdot \frac{y_1^2 - a^2}{1 + a^2} \]
Multiplying through by \(1 + a^2\) leads to:
\[ 2y_1x_1 = c(y_1^2 - a^2) \]

### Step 7: Rearranging to find the locus
Rearranging gives:
\[ cy_1^2 - 2y_1x_1 - ca^2 = 0 \]
This is a quadratic equation in \(y_1\).

### Step 8: Final form of the locus
To express this in standard form, we can replace \(x_1\) with \(x\) and \(y_1\) with \(y\):
\[ cy^2 - 2xy - ca^2 = 0 \]
This can be rearranged to:
\[ cy^2 - 2xy = ca^2 \]
Thus, the locus of point P is given by:
\[ cy^2 - 2xy - ca^2 = 0 \]