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 satisfies the given relation, let's first analyze the given equation. We are given:

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

Let's rewrite this to understand the structure of the equation.

1. **Complete the square** for both \(x\) and \(y\) terms:

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

   Completing the square for \(x\) and \(y\):

   \[
   (x - A)^2 + (y - B)^2 = A^2 + B^2 - C
   \]

   This equation represents a circle with center \((A, B)\) and radius \(\sqrt{A^2 + B^2 - C}\).

2. **Determine the Differential Equation**:

   To find a differential equation corresponding to this solution, differentiate both sides of the equation implicitly with respect to \(x\).

   Since \((x - A)^2 + (y - B)^2 = A^2 + B^2 - C\), we differentiate each side with respect to \(x\):

   \[
   2(x - A) + 2(y - B) \frac{dy}{dx} = 0
   \]

   Simplify by dividing by 2:

   \[
   (x - A) + (y - B) \frac{dy}{dx} = 0
   \]

3. **Rewrite the Resulting Differential Equation**:

   Rearranging terms, we obtain:

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

This is the required differential equation. The solution to this differential equation will be a family of circles centered at \((A, B)\) with radius \(\sqrt{A^2 + B^2 - C}\), under the condition that \(A^2 + B^2 > C\).

### Summary
The differential equation is:

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

Would you like further details or have any specific questions?

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### Related Questions
1. How do you find the general solution to this differential equation?
2. What are the possible values of \(A\), \(B\), and \(C\) if the radius is zero?
3. How does the differential equation change if we vary \(C\)?
4. What happens if \(A^2 + B^2 = C\)?
5. Can this differential equation represent any other geometric shapes?

### Tip
When dealing with equations of circles, completing the square can often simplify the process of finding a related differential equation.

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

Learn how to derive the differential equation of a family of circles represented by the equation x^2 + y^2 - 2Ax - 2By + C = 0. This detailed solution shows the steps of completing the square, implicit differentiation, and how to obtain the corresponding differential equation. The solution is valid under the condition A^2 + B^2 > C.

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