Find a differential equation whose solution Where A, B, C arbitrary constants and A ^ 2 + B ^ 2 > C x ^ 2 + y ^ 2 - 2Ax - 2By + C = 0



To find a differential equation whose solution matches the given expression:

\[
x^2 + y^2 - 2Ax - 2By + C = 0,
\]

we'll start by analyzing and differentiating this equation.

### Step 1: Analyze the Equation
The equation provided can be rewritten as:
\[
(x - A)^2 + (y - B)^2 = A^2 + B^2 - C.
\]
This represents a circle centered at \((A, B)\) with a radius \(\sqrt{A^2 + B^2 - C}\), given that \(A^2 + B^2 > C\).

### Step 2: Implicit Differentiation
To find a differential equation, differentiate the equation implicitly with respect to \(x\). Doing this will introduce \(y'\), helping us find a relationship between \(x\), \(y\), and \(y'\).

Starting with:
\[
x^2 + y^2 - 2Ax - 2By + C = 0,
\]
we take the derivative with respect to \(x\):
\[
2x + 2y \cdot \frac{dy}{dx} - 2A - 2B \cdot \frac{dy}{dx} = 0.
\]
Simplify this expression to isolate \(\frac{dy}{dx}\):
\[
(2y - 2B) \frac{dy}{dx} = 2A - 2x.
\]
Now we get:
\[
\frac{dy}{dx} = \frac{x - A}{y - B}.
\]

### Step 3: Second Derivative for a Differential Equation
We can take the derivative of \(\frac{dy}{dx} = \frac{x - A}{y - B}\) again with respect to \(x\) to involve \(y''\) for a second-order differential equation:

Differentiating \(\frac{dy}{dx} = \frac{x - A}{y - B}\) with respect to \(x\):
\[
\frac{d^2y}{dx^2} = \frac{(y - B) \cdot 1 - (x - A) \cdot \frac{dy}{dx}}{(y - B)^2}.
\]
Substituting \(\frac{dy}{dx} = \frac{x - A}{y - B}\) back into this expression gives us a differential equation that involves \(y\), \(y'\), and \(y''\).

This differential equation describes the family of circles centered at \((A, B)\) with radius \(\sqrt{A^2 + B^2 - C}\), consistent with the original equation.

Would you like more details on the second differentiation or have any questions?

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Here are 5 additional related questions for exploration:

1. How would we solve a differential equation that includes both \(y'\) and \(y''\)?
2. How can implicit differentiation be applied to other conic sections?
3. What other types of differential equations can represent geometric shapes?
4. How would we find a general solution for a second-order differential equation in polar coordinates?
5. How does the concept of orthogonal trajectories relate to families of circles?

**Tip:** Implicit differentiation is a powerful tool for deriving relationships between variables when one variable depends on another but is not explicitly solved for.

Find a differential equation whose solution is x^2 + y^2 - 2Ax - 2By + C = 0, where A, B, C are arbitrary constants and A^2 + B^2 > C.

Learn how to derive the differential equation for a family of circles represented by the equation x^2 + y^2 - 2Ax - 2By + C = 0. This process involves implicit differentiation, geometric analysis, and second-order differential equations. Suitable for students in Grades 11-12 studying calculus and geometry.

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