Differential equation of the family of circles touching the line `y=2` at `(0,2)` is

Options

A. `x^2 + (y-2)^2 + dy/dx (y-2) =0`
B. `x^2 +(y-2)[2-2xdx/dy-y]=0`
C. `x^2 + (y-2)^2 +(dx/dy +y -2)(y-2)=0`
D. none of these

Answer

To find the differential equation of the family of circles that touch the line \( y = 2 \) at the point \( (0, 2) \), we can follow these steps:

### Step 1: Understand the Geometry
The circles that touch the line \( y = 2 \) at the point \( (0, 2) \) will have their centers along the vertical line \( x = 0 \) (the y-axis) and will be positioned either above or below the line \( y = 2 \).

### Step 2: Define the Circle
Let the center of the circle be at the point \( (0, k) \) where \( k \) is the y-coordinate of the center. The radius \( r \) of the circle will be the distance from the center to the line \( y = 2 \).

### Step 3: Relate the Radius and Center
Since the circle touches the line \( y = 2 \), we can express the radius \( r \) in terms of \( k \):
- If the center is above the line, \( r = k - 2 \).
- If the center is below the line, \( r = 2 - k \).

For our case, since the circle can be either above or below the line, we can write:
\[ r = |k - 2| \]

### Step 4: Write the Equation of the Circle
The general equation of a circle with center \( (0, k) \) and radius \( r \) is given by:
\[ x^2 + (y - k)^2 = r^2 \]
Substituting \( r = |k - 2| \), we have:
\[ x^2 + (y - k)^2 = (k - 2)^2 \]

### Step 5: Expand and Rearrange
Expanding the equation:
\[ x^2 + (y^2 - 2ky + k^2) = (k^2 - 4k + 4) \]
This simplifies to:
\[ x^2 + y^2 - 2ky + k^2 = k^2 - 4k + 4 \]
Cancelling \( k^2 \) from both sides gives:
\[ x^2 + y^2 - 2ky = -4k + 4 \]
Rearranging:
\[ x^2 + y^2 + 4 = 2ky \]

### Step 6: Solve for \( k \)
From the equation, we can express \( k \):
\[ k = \frac{x^2 + y^2 + 4}{2y} \]

### Step 7: Differentiate to Eliminate \( k \)
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(k) = \frac{d}{dx}\left(\frac{x^2 + y^2 + 4}{2y}\right) \]
Using the quotient rule on the right-hand side:
\[ \frac{dk}{dx} = \frac{(2x + 2y \frac{dy}{dx}) \cdot 2y - (x^2 + y^2 + 4) \cdot 2 \frac{dy}{dx}}{(2y)^2} \]

### Step 8: Substitute \( k \) Back
Substituting \( k \) back into the differentiated equation will allow us to eliminate \( k \) and form a differential equation.

### Final Step: Formulate the Differential Equation
After substituting and simplifying, we arrive at the required differential equation, which can be expressed as:
\[ (x^2 + y^2 - 4) \frac{dy}{dx} = x + y \frac{dy}{dx} (2y - 2) \]