The degree of the differential equation representing the family of curves `(x-a)^(2)+y^(2)=16` is :

Options

A. 0
B. 2
C. 3
D. 1

Answer

To find the degree of the differential equation representing the family of curves given by \((x-a)^2 + y^2 = 16\), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation of the curve:
\[ (x-a)^2 + y^2 = 16 \]
We will differentiate both sides with respect to \(x\).

### Step 2: Apply differentiation
Differentiating the left-hand side:
\[ \frac{d}{dx}[(x-a)^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[16] \]
Using the chain rule:
\[ 2(x-a) \cdot \frac{d}{dx}(x-a) + 2y \cdot \frac{dy}{dx} = 0 \]
Since \(\frac{d}{dx}(x-a) = 1\), we have:
\[ 2(x-a) + 2y \frac{dy}{dx} = 0 \]

### Step 3: Simplify the equation
Dividing the entire equation by 2:
\[ (x-a) + y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ y \frac{dy}{dx} = -(x-a) \]

### Step 4: Express \(a\) in terms of \(x\) and \(y\)
From the rearranged equation, we can express \(a\):
\[ a = x + y \frac{dy}{dx} \]

### Step 5: Substitute \(a\) back into the original equation
Now, substituting \(a\) back into the original equation:
\[ (x - (x + y \frac{dy}{dx}))^2 + y^2 = 16 \]
This simplifies to:
\[ (-y \frac{dy}{dx})^2 + y^2 = 16 \]
Expanding gives:
\[ y^2 \left(\frac{dy}{dx}\right)^2 + y^2 = 16 \]
Factoring out \(y^2\):
\[ y^2 \left(\left(\frac{dy}{dx}\right)^2 + 1\right) = 16 \]

### Step 6: Form the differential equation
We can express the differential equation as:
\[ y^2 \left(\frac{dy}{dx}\right)^2 + y^2 - 16 = 0 \]

### Step 7: Identify the degree of the differential equation
The degree of a differential equation is defined as the power of the highest order derivative in the equation. Here, the highest order derivative is \(\frac{dy}{dx}\) and it appears squared.

Thus, the degree of the differential equation is:
\[ \text{Degree} = 2 \]

### Final Answer:
The degree of the differential equation representing the family of curves \((x-a)^2 + y^2 = 16\) is **2**.

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