The differential equation of family of circles whose centre lies on x-axis, is

Options

A. `(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)+1=0`
B. `y(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)-1=0`
C. `y(d^(2)y)/(dx^(2))-((dy)/(dx))^(2)-1=0`
D. `y(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)+1=0`

Answer

To find the differential equation of the family of circles whose centers lie on the x-axis, we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center at \((h, 0)\) and radius \(R\) is given by:
\[ (x - h)^2 + y^2 = R^2 \]

### Step 2: Differentiate the equation with respect to \(x\)
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}[(x - h)^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[R^2] \]
This results in:
\[ 2(x - h) + 2y \frac{dy}{dx} = 0 \]
Simplifying this gives:
\[ (x - h) + y \frac{dy}{dx} = 0 \]

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

### Step 4: Differentiate again
Now we differentiate the equation \(h = x + y \frac{dy}{dx}\) with respect to \(x\):
\[ \frac{dh}{dx} = 1 + \left(\frac{dy}{dx} + y \frac{d^2y}{dx^2}\right) \]
Since \(h\) is a constant (it represents the center of the circle), its derivative is zero:
\[ 0 = 1 + \frac{dy}{dx} + y \frac{d^2y}{dx^2} \]

### Step 5: Rearranging the equation
Rearranging the above equation gives us the final form of the differential equation:
\[ y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 1 = 0 \]

### Conclusion
Thus, the differential equation of the family of circles whose centers lie on the x-axis is:
\[ y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 1 = 0 \]

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