To find a differential equation whose solution is the equation $x^2 + y^2 - 2Ax - 2By + C = 0$, we will differentiate the equation with respect to $x$ and $y$, eliminating the arbitrary constants $A$, $B$, and $C$.

Step 1: Start with the given equation:

$x^2 + y^2 - 2Ax - 2By + C = 0$ Here, $A$, $B$, and $C$ are arbitrary constants.

Step 2: Differentiate the equation with respect to $x$.

Differentiate both sides of the equation implicitly with respect to $x$:

$\frac{d}{dx}(x^2 + y^2 - 2Ax - 2By + C) = 0$

Using the chain rule:

$2x + 2y \frac{dy}{dx} - 2A - 2B \frac{dy}{dx} = 0$

This simplifies to:

$2x - 2A + 2y \frac{dy}{dx} - 2B \frac{dy}{dx} = 0$

Now, divide through by 2:

$x - A + y \frac{dy}{dx} - B \frac{dy}{dx} = 0$

Rearrange this to isolate the terms involving $\frac{dy}{dx}$:

$x - A = (B - y) \frac{dy}{dx}$

Thus, we have:

$\frac{dy}{dx} = \frac{x - A}{B - y}$

Step 3: Differentiate with respect to $y$.

Next, differentiate the original equation implicitly with respect to $y$:

$\frac{d}{dy}(x^2 + y^2 - 2Ax - 2By + C) = 0$

Applying the chain rule:

$2y - 2B \frac{dx}{dy} = 0$

Simplifying:

$y - B \frac{dx}{dy} = 0$

Thus:

$\frac{dx}{dy} = \frac{y}{B}$

Step 4: Substitute $\frac{dx}{dy}$ into the previous equation.

To eliminate the arbitrary constants $A$, $B$, and $C$, we can substitute $\frac{dx}{dy} = \frac{y}{B}$ into the differential equation we derived earlier.

Now we substitute:

$\frac{dy}{dx} = \frac{x - A}{B - y}$

Thus, the final differential equation is:

$\frac{dy}{dx} = \frac{x - A}{B - y}$

Where we replaced $A$ and $B$ using the relationships established.

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Let me know if you'd like any further clarification or more detailed steps!

Additional Questions:

1. How would you approach solving this differential equation?
2. What kind of behavior does the solution $x^2 + y^2 - 2Ax - 2By + C = 0$ describe geometrically?
3. How would you modify this differential equation to fit an ellipse or a circle?
4. What are the steps involved in solving a first-order linear differential equation?
5. How can this method be generalized for higher-order differential equations?

Tip: When differentiating implicitly, always be mindful of applying the chain rule and properly handling terms involving both $x$ and $y$.