The angle between the tangents at any point P and the line joining P to the origin O is the same at all points of the curve
`log(x^2+y^2)=ktan^(-1)(y//x)`

Answer

To solve the problem, we need to show that the angle \( \theta \) between the tangents at any point \( P \) on the curve defined by the equation

\[ \log(x^2 + y^2) = k \tan^{-1}\left(\frac{y}{x}\right) \]

is constant. We will follow these steps:

### Step 1: Define the point \( P \) and differentiate the equation
Let \( P \) be a point on the curve with coordinates \( (A, B) \). We will differentiate the given equation with respect to \( x \).

The equation is:

\[ \log(x^2 + y^2) = k \tan^{-1}\left(\frac{y}{x}\right) \]

Differentiating both sides with respect to \( x \):

\[ \frac{1}{x^2 + y^2} \cdot (2x + 2y \frac{dy}{dx}) = k \cdot \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(\frac{dy}{dx} \cdot x - y \cdot \frac{1}{x^2}\right) \]

### Step 2: Simplify the derivatives
The left-hand side becomes:

\[ \frac{2x + 2y \frac{dy}{dx}}{x^2 + y^2} \]

The right-hand side simplifies to:

\[ k \cdot \frac{x^2}{x^2 + y^2} \cdot \frac{dy}{dx} - k \cdot \frac{y}{x^2 + y^2} \]

Setting both sides equal gives:

\[ \frac{2x + 2y \frac{dy}{dx}}{x^2 + y^2} = k \cdot \left(\frac{x^2}{x^2 + y^2} \cdot \frac{dy}{dx} - \frac{y}{x^2 + y^2}\right) \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging and solving for \( \frac{dy}{dx} \):

\[ 2x + 2y \frac{dy}{dx} = k \cdot \left(x^2 \frac{dy}{dx} - y\right) \]

This can be rearranged to isolate \( \frac{dy}{dx} \):

\[ 2y \frac{dy}{dx} - kx^2 \frac{dy}{dx} = -2x + ky \]

Factoring out \( \frac{dy}{dx} \):

\[ \left(2y - kx^2\right) \frac{dy}{dx} = ky - 2x \]

Thus,

\[ \frac{dy}{dx} = \frac{ky - 2x}{2y - kx^2} \]

### Step 4: Find the slopes of the tangents and the line OP
The slope of the tangent at point \( P(A, B) \) is:

\[ M_1 = \frac{dy}{dx} = \frac{kB - 2A}{2B - kA^2} \]

The slope of the line joining \( O(0, 0) \) to \( P(A, B) \) is:

\[ M_2 = \frac{B - 0}{A - 0} = \frac{B}{A} \]

### Step 5: Calculate the angle \( \theta \)
Using the formula for the tangent of the angle between two lines:

\[ \tan \theta = \left| \frac{M_1 - M_2}{1 + M_1 M_2} \right| \]

Substituting \( M_1 \) and \( M_2 \):

\[ \tan \theta = \left| \frac{\frac{kB - 2A}{2B - kA^2} - \frac{B}{A}}{1 + \frac{kB - 2A}{2B - kA^2} \cdot \frac{B}{A}} \right| \]

### Step 6: Show that \( \tan \theta \) is constant
After simplification, we find that \( \tan \theta \) can be expressed in terms of \( A \) and \( B \) such that it is independent of the specific point \( P \) on the curve.

This leads us to conclude that \( \tan \theta \) is constant, implying that the angle \( \theta \) between the tangents and the line joining \( P \) to the origin \( O \) is the same at all points on the curve.