derive a differential equation for the family of circles (x-lambda)^2 + y^2 = lambda ^2 (lambda >0)

To derive the differential equation for the given family of circles:

\[
(x - \lambda)^2 + y^2 = \lambda^2, \quad \text{where } \lambda > 0,
\]

we will follow these steps:

1. Treat \(\lambda\) as a parameter.
2. Differentiate with respect to \(x\) to obtain a relationship involving \(y\), \(\frac{dy}{dx}\), and \(\lambda\).
3. Eliminate the parameter \(\lambda\) to get a differential equation that relates only \(x\), \(y\), and \(\frac{dy}{dx}\).

### Step 1: Original Equation

The family of circles can be written as:

\[
(x - \lambda)^2 + y^2 = \lambda^2.
\]

### Step 2: Differentiate with Respect to \(x\)

Differentiate both sides with respect to \(x\):

\[
\frac{d}{dx} \left[(x - \lambda)^2 + y^2 \right] = \frac{d}{dx} [\lambda^2].
\]

This gives:

\[
2(x - \lambda) \cdot \frac{d}{dx} (x - \lambda) + 2y \frac{dy}{dx} = 0.
\]

Since \(\lambda\) is a constant with respect to \(x\), \(\frac{d}{dx} (x - \lambda) = 1\), so:

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

Simplify to:

\[
x - \lambda + y \frac{dy}{dx} = 0.
\]

### Step 3: Solve for the Parameter \(\lambda\)

From the original equation:

\[
(x - \lambda)^2 = \lambda^2 - y^2.
\]

Taking the square root:

\[
x - \lambda = \pm \sqrt{\lambda^2 - y^2}.
\]

Solve for \(\lambda\):

\[
\lambda = x \mp \sqrt{\lambda^2 - y^2}.
\]

Square both sides:

\[
\lambda^2 = (x \mp \sqrt{\lambda^2 - y^2})^2.
\]

This expands to:

\[
\lambda^2 = x^2 \pm 2x \sqrt{\lambda^2 - y^2} + (\lambda^2 - y^2).
\]

Rearrange terms:

\[
0 = x^2 - y^2 \pm 2x \sqrt{\lambda^2 - y^2}.
\]

Solve for \(\sqrt{\lambda^2 - y^2}\):

\[
\sqrt{\lambda^2 - y^2} = \frac{y^2 - x^2}{2x}.
\]

Substitute back into the expression for \(x - \lambda\):

\[
x - \lambda = \frac{y^2 - x^2}{2x}.
\]

Thus:

\[
\lambda = x - \frac{y^2 - x^2}{2x}.
\]

### Step 4: Substitute \(\lambda\) in the Differential Equation

Now substitute \(\lambda = x - \frac{y^2 - x^2}{2x}\) in the differential equation:

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

Simplify:

\[
\frac{y^2 - x^2}{2x} + y \frac{dy}{dx} = 0.
\]

Multiply by \(2x\):

\[
y^2 - x^2 + 2xy \frac{dy}{dx} = 0.
\]

### Final Differential Equation

The differential equation for the family of circles is:

\[
2xy \frac{dy}{dx} = x^2 - y^2.
\]

Or, equivalently:

\[
\frac{dy}{dx} = \frac{x^2 - y^2}{2xy}.
\]

### Summary

The derived differential equation for the family of circles \((x - \lambda)^2 + y^2 = \lambda^2\) is:

\[
\frac{dy}{dx} = \frac{x^2 - y^2}{2xy}.
\]

Would you like more details or have any questions?

#### Related Questions

1. What are the solutions to this differential equation?
2. How can we verify if a particular curve is a solution to this differential equation?
3. Can this method be generalized to other families of curves, like ellipses or hyperbolas?
4. What is the geometric interpretation of the derived differential equation?
5. How does this differential equation change if the parameter \(\lambda\) can take negative values?

**Tip:** When dealing with a family of curves, identifying a parameter that characterizes the family is crucial for eliminating it to derive the differential equation.

Derive a differential equation for the family of circles (x-lambda)^2 + y^2 = lambda^2 (lambda > 0).

This article provides a step-by-step guide to deriving the differential equation for the family of circles (x - λ)^2 + y^2 = λ^2. Learn how to use implicit differentiation and parameter elimination to find the equation dy/dx = (x^2 - y^2) / (2xy). Suitable for advanced high school or undergraduate calculus students.

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