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We are tasked with finding the value of \( r \) for the tangent line to the equation 

\[
x^2 + y^2 = (3x^2 + 8y^2 - 5x)^2
\]

at the point \( \left(0, \frac{1}{8}\right) \), where the tangent line is given by:

\[
y = \frac{1}{8} + r(x - 0).
\]

### Steps:

1. **Differentiate both sides of the given equation implicitly** with respect to \( x \). This will help us find the slope \( r \) of the tangent line.

    Given equation: 
    \[
    x^2 + y^2 = \left(3x^2 + 8y^2 - 5x\right)^2
    \]

2. Use implicit differentiation on both sides.

   - The derivative of the left-hand side \( x^2 + y^2 \) is:
     \[
     \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 2x + 2y \frac{dy}{dx}
     \]

   - The derivative of the right-hand side \( \left(3x^2 + 8y^2 - 5x\right)^2 \) will require the chain rule:
     \[
     \frac{d}{dx} \left(3x^2 + 8y^2 - 5x\right)^2 = 2 \cdot \left(3x^2 + 8y^2 - 5x\right) \cdot \frac{d}{dx} \left(3x^2 + 8y^2 - 5x\right)
     \]
     
     Now, differentiate \( 3x^2 + 8y^2 - 5x \):
     \[
     \frac{d}{dx} \left(3x^2 + 8y^2 - 5x\right) = 6x + 16y \frac{dy}{dx} - 5
     \]

3. After differentiation, substitute the point \( \left(0, \frac{1}{8}\right) \) into the resulting equation to solve for \( \frac{dy}{dx} \), which will give the value of \( r \).

I'll work out the details for this differentiation step-by-step.

### Differentiating and Substituting

#### Left-hand side:

\[
2x + 2y \frac{dy}{dx}
\]

At \( x = 0 \) and \( y = \frac{1}{8} \):

\[
2(0) + 2 \left(\frac{1}{8}\right) \frac{dy}{dx} = \frac{1}{4} \frac{dy}{dx}
\]

#### Right-hand side:

\[
2 \cdot (3x^2 + 8y^2 - 5x) \cdot (6x + 16y \frac{dy}{dx} - 5)
\]

At \( x = 0 \) and \( y = \frac{1}{8} \), we substitute into \( (3x^2 + 8y^2 - 5x) \) and find:

\[
3(0)^2 + 8\left(\frac{1}{8}\right)^2 - 5(0) = \frac{8}{64} = \frac{1}{8}.
\]

Now, substitute into the derivative term \( 6x + 16y \frac{dy}{dx} - 5 \):

\[
6(0) + 16\left(\frac{1}{8}\right) \frac{dy}{dx} - 5 = 2 \frac{dy}{dx} - 5.
\]

Thus, the right-hand side becomes:

\[
2 \cdot \frac{1}{8} \cdot \left(2 \frac{dy}{dx} - 5\right) = \frac{1}{4} \left(2 \frac{dy}{dx} - 5\right).
\]

### Equating Both Sides:

We now equate the two sides:

\[
\frac{1}{4} \frac{dy}{dx} = \frac{1}{4} \left(2 \frac{dy}{dx} - 5\right).
\]

Simplifying:

\[
\frac{dy}{dx} = 2 \frac{dy}{dx} - 5
\]

\[
0 = \frac{dy}{dx} - 5
\]

\[
\frac{dy}{dx} = 5.
\]

Thus, the slope of the tangent line is \( r = 5 \).

### Final Answer:

The value of \( r \) is \( \boxed{5} \).

---

Would you like more details or have any questions? Here are some related questions for further exploration:

1. What is the geometric interpretation of the slope of a tangent line?
2. How does implicit differentiation differ from explicit differentiation?
3. Can the chain rule be applied to functions other than polynomials? How?
4. What are some real-life applications of finding tangent lines to curves?
5. How would the process change if the equation had been parametric?

**Tip:** Always double-check substitution points in differentiation to ensure accuracy in complex expressions.

The function of the tangent line to x^2 + y^2 = (3x^2 + 8y^2 - 5x)^2 at the point (0, 1/8) is defined as y = 1/8 + r(x - 0). Determine the value for r.

Learn how to use implicit differentiation and the chain rule to find the slope (r) of the tangent line to the equation x^2 + y^2 = (3x^2 + 8y^2 - 5x)^2 at the point (0, 1/8). This detailed step-by-step solution covers key calculus concepts and is suitable for advanced high school students in Grades 11-12.

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