The differential equation satisfied by all the circles with radius = r(constant) and center on the line x=y is

Options

A. `(x-y)^2 (1+y_(1) ^2)= r^2 (1 + y_(1))^(2)`
B. `(x-y) y_(1)=r^2 (1 + y_(1)) (1 + y_(1))^(2)`
C. `(x-y) y_(1)^(2) =r^2 (1 + y_(1)^(2))y_(2)`
D. `(x-y) y_(1)^(2)= r^2(1 + y_(1))y_(2)^(2)`

Answer

To find the differential equation satisfied by all the circles with radius \( r \) (a constant) and center on the line \( x = y \), we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center \( (h, h) \) (since the center lies on the line \( x = y \)) and radius \( r \) is given by:
\[ (x - h)^2 + (y - h)^2 = r^2 \]
This is our equation (1).

### Step 2: Differentiate the equation
We will differentiate equation (1) with respect to \( x \) to eliminate the parameter \( h \). Using implicit differentiation:
\[ \frac{d}{dx}[(x - h)^2] + \frac{d}{dx}[(y - h)^2] = \frac{d}{dx}[r^2] \]
This gives us:
\[ 2(x - h)(1) + 2(y - h)\frac{dy}{dx} = 0 \]
Simplifying this, we have:
\[ (x - h) + (y - h)\frac{dy}{dx} = 0 \]

### Step 3: Solve for \( h \)
Rearranging the equation, we find:
\[ h = x + (y - h)\frac{dy}{dx} \]
From this, we can isolate \( h \):
\[ h = \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}} \]

### Step 4: Substitute \( h \) back into equation (1)
Now we substitute the value of \( h \) back into the original circle equation:
\[ \left(x - \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\right)^2 + \left(y - \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\right)^2 = r^2 \]

### Step 5: Simplify the equation
After substituting and simplifying, we will arrive at a form that can be expressed in terms of \( \frac{dy}{dx} \) and \( r \).

### Step 6: Final form of the differential equation
After careful simplification, we will find that the differential equation can be expressed as:
\[ (x - y)^2(1 + \left(\frac{dy}{dx}\right)^2) = r^2(1 + \left(\frac{dy}{dx}\right)^2) \]

### Conclusion
Thus, the differential equation satisfied by all the circles with radius \( r \) and center on the line \( x = y \) is:
\[ (x - y)^2(1 + (y')^2) = r^2(1 + (y')^2) \]
where \( y' = \frac{dy}{dx} \).