Let's start by analyzing and rearranging the given equation:

\[
x^2 + y^2 - 4x - 14 = 0
\]

To express this equation in the standard form of a circle \((x - h)^2 + (y - k)^2 = r^2\), we need to complete the square for the \(x\) terms and simplify the equation.

### Step 1: Rearrange the equation
Move the constant term to the other side:

\[
x^2 + y^2 - 4x = 14
\]

### Step 2: Complete the square for the \(x\) terms

For the \(x\) terms:

\[
x^2 - 4x
\]

To complete the square, take half of the coefficient of \(x\), which is \(-4\), divide by 2 to get \(-2\), and then square it to get \(4\).

So, add and subtract 4 inside the equation:

\[
(x^2 - 4x + 4) + y^2 = 14 + 4
\]

This simplifies to:

\[
(x - 2)^2 + y^2 = 18
\]

### Step 3: Interpret the equation

Now, the equation is in the standard form of a circle:

\[
(x - 2)^2 + y^2 = 18
\]

- **Center**: The center of the circle is at \((2, 0)\).
- **Radius**: The radius \(r\) is the square root of 18, which simplifies to \(r = \sqrt{18} = 3\sqrt{2}\).

### Summary:
- **Center**: \((2, 0)\)
- **Radius**: \(3\sqrt{2}\)

Would you like more details or have any further questions?

Here are some related questions:

1. How do you graph this circle?
2. How do you complete the square for a quadratic equation?
3. What is the area of this circle?
4. How do you find the circumference of the circle?
5. How does the equation change if the center of the circle is moved to \((0, 0)\)?
6. What if the radius were doubled? How would the equation look?
7. How does completing the square help in converting equations to standard forms?
8. What are the key differences between this equation and the general equation of a circle?

**Tip:** Completing the square is a valuable technique for transforming quadratic equations into more useful forms, such as the standard form of a circle.

Learn how to solve the equation x² + y² - 4x - 14 = 0 using the standard form of a circle equation. This problem involves completing the square for quadratic equations and interpreting the results geometrically. Suitable for students in Grades 10-12.

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