Form differential equation for `(x-a)^(2)+y^(2)=a^(2)` A)`(x+yy_(1))^(2)=y^(2)` B)`(xy_(1)+y)^(2)=y^(2)` C)`y^(2)=x^(2)+2xyy_(1)` D)`x^(2)=y^(2)+2xyy_(1)`

Options

A. `(x+yy_(1))^(2)=y^(2)`
B. `(xy_(1)+y)^(2)=y^(2)`
C. `y^(2)=x^(2)+2xyy_(1)`
D. `x^(2)=y^(2)+2xyy_(1)`

Answer

To form the differential equation from the given equation \((x-a)^2 + y^2 = a^2\), we will follow these steps:

### Step 1: Expand the given equation
Start with the equation:
\[ (x-a)^2 + y^2 = a^2 \]
Using the identity \((a-b)^2 = a^2 - 2ab + b^2\), we expand:
\[ x^2 - 2ax + a^2 + y^2 = a^2 \]
Now, simplify by canceling \(a^2\) from both sides:
\[ x^2 - 2ax + y^2 = 0 \]

### Step 2: Differentiate the equation
Now we differentiate the equation with respect to \(x\):
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(2ax) + \frac{d}{dx}(y^2) = 0 \]
This gives:
\[ 2x - 2a + 2y \frac{dy}{dx} = 0 \]
We can express \(\frac{dy}{dx}\) as \(y'\):
\[ 2x - 2a + 2y y' = 0 \]

### Step 3: Solve for \(a\)
From the differentiated equation, we can isolate \(a\):
\[ 2a = 2x + 2y y' \]
Thus, we have:
\[ a = x + y y' \]

### Step 4: Substitute \(a\) back into the original equation
Now substitute \(a\) back into the original equation:
\[ x^2 + y^2 - 2a x = 0 \]
Substituting \(a\):
\[ x^2 + y^2 - 2(x + y y')x = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x^2 - 2xy y' = 0 \]
Rearranging gives:
\[ y^2 = x^2 + 2xy y' \]

### Final Result
The final differential equation is:
\[ y^2 = x^2 + 2xy y' \]