Key Concepts and Formulas

• Equation of a Circle: The general equation of a circle with center $(h, k)$ and radius $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$.
• Differential Equation Formation: To form a differential equation from a family of curves, differentiate the equation of the family as many times as the number of arbitrary constants, and then eliminate the arbitrary constants from the equations obtained.

Step-by-Step Solution

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

Since the circle passes through the origin and its center lies on the line $y = x$, let the center be $(a, a)$. The radius of the circle is then the distance from the center to the origin, which is $\sqrt{a^2 + a^2} = \sqrt{2a^2} = |a|\sqrt{2}$. The equation of the circle is: $(x - a)^2 + (y - a)^2 = (\sqrt{2}a)^2$ Expanding and simplifying, we get: $x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = 2a^2$ $x^2 + y^2 - 2ax - 2ay = 0$ This equation has one arbitrary constant, $a$.

Step 2: Differentiate with Respect to $x$

Differentiate the equation $x^2 + y^2 - 2ax - 2ay = 0$ with respect to $x$: $\frac{d}{dx}(x^2 + y^2 - 2ax - 2ay) = \frac{d}{dx}(0)$ $2x + 2y\frac{dy}{dx} - 2a - 2a\frac{dy}{dx} = 0$ $x + y\frac{dy}{dx} - a - a\frac{dy}{dx} = 0$ $a(1 + \frac{dy}{dx}) = x + y\frac{dy}{dx}$ $a = \frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}$

Step 3: Eliminate the Arbitrary Constant $a$

Substitute the value of $a$ obtained in Step 2 back into the equation of the circle: $x^2 + y^2 - 2x\left(\frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\right) - 2y\left(\frac{x + y\frac{dy}{dx}}{1 + \frac{dy}{dx}}\right) = 0$ Multiply both sides by $(1 + \frac{dy}{dx})$: $(x^2 + y^2)(1 + \frac{dy}{dx}) - 2x(x + y\frac{dy}{dx}) - 2y(x + y\frac{dy}{dx}) = 0$ $x^2 + y^2 + (x^2 + y^2)\frac{dy}{dx} - 2x^2 - 2xy\frac{dy}{dx} - 2xy - 2y^2\frac{dy}{dx} = 0$ $-x^2 - 2xy + y^2 + (x^2 - 2xy - y^2)\frac{dy}{dx} = 0$ $(x^2 - 2xy - y^2)\frac{dy}{dx} = x^2 + 2xy - y^2$ $(x^2 - y^2 - 2xy)\frac{dy}{dx} = x^2 - y^2 + 2xy$ $(x^2 - y^2 - 2xy)dy = (x^2 - y^2 + 2xy)dx$ $(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy$ $(x^2 - y^2 + 2xy)dx - (x^2 - y^2 - 2xy)dy = 0$

Common Mistakes & Tips

• Careful Differentiation: Ensure correct application of the chain rule when differentiating implicitly.
• Algebraic Manipulation: Pay close attention to signs and distribution during the elimination process.
• Checking the Answer: After obtaining the differential equation, verify if it matches any of the given options.

Summary

We started with the general equation of a family of circles passing through the origin and having centers on the line $y = x$. After differentiating this equation with respect to $x$ and eliminating the arbitrary constant, we arrived at the differential equation $(x^2 - y^2 + 2xy)dx = (x^2 - y^2 - 2xy)dy$. This matches option (A).

Final Answer

The final answer is $\boxed{\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y}$, which corresponds to option (A).