Key Concepts and Formulas

• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$.
• Differential Equations: A differential equation is an equation involving derivatives of a function. The order of the differential equation is the highest order derivative appearing in the equation.
• Chain Rule: The chain rule is used to differentiate composite functions. For example, $\frac{d}{dx} f(y) = \frac{df}{dy} \cdot \frac{dy}{dx}$.

Step-by-Step Solution

Step 1: Establishing the General Equation of the Family of Circles

We are given that the circles pass through the origin and have their centers on the $x$-axis. Let the center of the circle be $(h, 0)$ and the radius be $r$. The general equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$. Since the center lies on the $x$-axis, $k = 0$. Thus, the equation becomes:

$(x-h)^2 + y^2 = r^2 \quad \ldots(1)$

Since the circle passes through the origin $(0, 0)$, we can substitute $x = 0$ and $y = 0$ into equation (1):

$(0-h)^2 + 0^2 = r^2$ $h^2 = r^2 \quad \ldots(2)$

Now, substitute $r^2 = h^2$ back into equation (1):

$(x-h)^2 + y^2 = h^2$ $x^2 - 2hx + h^2 + y^2 = h^2$ $x^2 + y^2 - 2hx = 0 \quad \ldots(A)$

This equation represents the family of circles passing through the origin and having their centers on the $x$-axis. The only arbitrary constant is $h$.

Step 2: Differentiating to Eliminate the Arbitrary Constant

Since there is one arbitrary constant, we need to differentiate equation (A) once with respect to $x$ to eliminate $h$.

Differentiating equation (A) with respect to $x$:

$\frac{d}{dx}(x^2 + y^2 - 2hx) = \frac{d}{dx}(0)$ $2x + 2y \frac{dy}{dx} - 2h = 0 \quad \ldots(3)$

Here, we used the chain rule to differentiate $y^2$ with respect to $x$: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.

Step 3: Eliminating the Arbitrary Constant

We now have two equations:

1. $x^2 + y^2 - 2hx = 0 \quad \ldots(A)$
2. $2x + 2y \frac{dy}{dx} - 2h = 0 \quad \ldots(3)$

We need to eliminate $h$. From equation (3), we can solve for $h$:

$2h = 2x + 2y \frac{dy}{dx}$ $h = x + y \frac{dy}{dx} \quad \ldots(4)$

Substitute this expression for $h$ into equation (A):

$x^2 + y^2 - 2x \left(x + y \frac{dy}{dx}\right) = 0$

Step 4: Simplifying to the Final Differential Equation

Simplify the equation:

$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$ $-x^2 + y^2 - 2xy \frac{dy}{dx} = 0$ $y^2 = x^2 + 2xy \frac{dy}{dx}$

This is the required differential equation.

Common Mistakes & Tips

• Remember to use the chain rule when differentiating $y$ terms with respect to $x$.
• Be careful with algebraic manipulations, especially when distributing and simplifying.
• Double-check the number of arbitrary constants to ensure you differentiate the correct number of times.

Summary

We found the differential equation of all circles passing through the origin and having their centers on the $x$-axis. We started with the general equation of a circle, applied the given conditions to find the specific form for this family, and then differentiated and eliminated the arbitrary constant to arrive at the differential equation ${y^2} = {x^2} + 2xy{{dy} \over {dx}}$.

The final answer is \boxed{{y^2} = {x^2} + 2xy{{dy} \over {dx}}}, which corresponds to option (A).