The differential equation of the family of circles with fixed radius 5 units and centre on the line `y=2` is

Options

A. `(y-2) y'^(2)=25-(y+2)^(2)`
B. `(y-2)^(2) y'^(2) =25 -(y-2)^(2)`
C. `(x-2)^(2) y'^(2) =25-(y-2)^(2)`
D. `(x-2)y'^(2)=25-(y+2)^(2)`

Answer

To find the differential equation of the family of circles with a fixed radius of 5 units and centers on the line \( y = 2 \), we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, the radius \(r = 5\) and the center's y-coordinate \(k = 2\) (since the center lies on the line \(y = 2\)). Thus, we can write the equation as:
\[ (x - h)^2 + (y - 2)^2 = 5^2 \]
This simplifies to:
\[ (x - h)^2 + (y - 2)^2 = 25 \]

### Step 2: Differentiate the equation
Next, we differentiate both sides of the equation with respect to \(x\):
\[ \frac{d}{dx}[(x - h)^2] + \frac{d}{dx}[(y - 2)^2] = \frac{d}{dx}[25] \]
Using the chain rule, we get:
\[ 2(x - h)(1) + 2(y - 2)\frac{dy}{dx} = 0 \]
This simplifies to:
\[ 2(x - h) + 2(y - 2)\frac{dy}{dx} = 0 \]
Dividing through by 2:
\[ (x - h) + (y - 2)\frac{dy}{dx} = 0 \]

### Step 3: Solve for \(x - h\)
From the equation, we can express \(x - h\) in terms of \(y\) and \(\frac{dy}{dx}\):
\[ x - h = - (y - 2)\frac{dy}{dx} \]

### Step 4: Substitute back into the original circle equation
Now, we substitute \(x - h\) back into the original circle equation:
\[ (- (y - 2)\frac{dy}{dx})^2 + (y - 2)^2 = 25 \]
This gives:
\[ (y - 2)^2 \left(\frac{dy}{dx}\right)^2 + (y - 2)^2 = 25 \]
Factoring out \((y - 2)^2\):
\[ (y - 2)^2 \left(\left(\frac{dy}{dx}\right)^2 + 1\right) = 25 \]

### Step 5: Rearranging the equation
Now, we can rearrange this equation:
\[ (y - 2)^2 \left(\frac{dy}{dx}\right)^2 = 25 - (y - 2)^2 \]

### Final Differential Equation
Thus, the differential equation of the family of circles is:
\[ (y - 2)^2 \left(\frac{dy}{dx}\right)^2 = 25 - (y - 2)^2 \]