STATEMENT-1 : The orthogonal trajectory of a family of circles touching x-axis at origin and whose centre the on y-axis is self orthogonal.
and
STATEMENT-2 : In order to find the orthogonal trajectory of a family of curves we put `-(dx)/(dy)` in place of `(dy)/(dx)` in the differential equation of the given family of curves.

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is true, Statement-2 is true, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem, we need to analyze the statements provided and derive the necessary conclusions step by step.

### Step 1: Understanding the Family of Circles
The family of circles mentioned in Statement 1 touches the x-axis at the origin and has centers on the y-axis. The general equation of such a circle can be written as:
\[ x^2 + (y - k)^2 = k^2 \]
where \( k \) is the y-coordinate of the center of the circle.

### Step 2: Simplifying the Circle Equation
Expanding the equation gives:
\[ x^2 + (y^2 - 2ky + k^2) = k^2 \]
This simplifies to:
\[ x^2 + y^2 - 2ky = 0 \]

### Step 3: Finding the Differential Equation
To find the orthogonal trajectory, we need to differentiate the equation with respect to \( x \):
\[ \frac{d}{dx}(x^2 + y^2 - 2ky) = 0 \]
This leads to:
\[ 2x + 2y \frac{dy}{dx} - 2k \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (2y - 2k) \frac{dy}{dx} = -2x \]
Thus,
\[ \frac{dy}{dx} = \frac{-2x}{2y - 2k} = \frac{-x}{y - k} \]

### Step 4: Finding the Orthogonal Trajectory
For orthogonal trajectories, we replace \(\frac{dy}{dx}\) with \(-\frac{dx}{dy}\):
\[ -\frac{dx}{dy} = \frac{-x}{y - k} \implies \frac{dx}{dy} = \frac{x}{y - k} \]

### Step 5: Solving the Differential Equation
This equation can be solved using separation of variables:
\[ \frac{dx}{x} = \frac{dy}{y - k} \]
Integrating both sides:
\[ \ln |x| = \ln |y - k| + C \]
Exponentiating gives:
\[ |x| = C|y - k| \]
Thus, we can express this as:
\[ x = C(y - k) \]

### Step 6: Analyzing Self-Orthogonality
To check if the orthogonal trajectories are self-orthogonal, we need to see if the resulting curves intersect orthogonally. This requires examining the slopes of the curves at their intersection points.

### Conclusion on Statements
1. **Statement 1**: The orthogonal trajectory of the family of circles is self-orthogonal. This is **false** because the derived orthogonal trajectories do not intersect the original family of circles orthogonally.
2. **Statement 2**: The method for finding orthogonal trajectories by replacing \(\frac{dy}{dx}\) with \(-\frac{dx}{dy}\) is indeed correct. This is **true**.

### Final Answer
Thus, Statement 1 is false, and Statement 2 is true.