Difference of a equation of a circle and a line by using a problem

Certainly! Let's go through an example problem that involves both the equation of a circle and the equation of a line. We'll then find their points of intersection, which will illustrate the difference between the two equations.

### Problem Example:
**Find the points of intersection between the circle \(x^2 + y^2 = 25\) and the line \(y = 3x + 1\).**

#### Step-by-Step Solution:

1. **Equation of the Circle:**
   \[
   x^2 + y^2 = 25
   \]
   This represents a circle with a center at the origin (0,0) and a radius of 5.

2. **Equation of the Line:**
   \[
   y = 3x + 1
   \]
   This represents a line with a slope of 3 and a y-intercept of 1.

3. **Substitute the Line Equation into the Circle Equation:**
   To find the points of intersection, we need to substitute \(y\) from the line equation into the circle equation:
   \[
   x^2 + (3x + 1)^2 = 25
   \]

4. **Expand and Simplify:**
   \[
   x^2 + (3x + 1)^2 = 25
   \]
   \[
   x^2 + (9x^2 + 6x + 1) = 25
   \]
   \[
   x^2 + 9x^2 + 6x + 1 = 25
   \]
   \[
   10x^2 + 6x + 1 = 25
   \]
   \[
   10x^2 + 6x + 1 - 25 = 0
   \]
   \[
   10x^2 + 6x - 24 = 0
   \]

5. **Solve the Quadratic Equation:**
   Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 10\), \(b = 6\), and \(c = -24\):
   \[
   x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 10 \cdot (-24)}}{2 \cdot 10}
   \]
   \[
   x = \frac{-6 \pm \sqrt{36 + 960}}{20}
   \]
   \[
   x = \frac{-6 \pm \sqrt{996}}{20}
   \]
   \[
   x = \frac{-6 \pm 2\sqrt{249}}{20}
   \]
   \[
   x = \frac{-3 \pm \sqrt{249}}{10}
   \]

6. **Find the Corresponding \(y\) Values:**
   Substitute \(x\) back into the line equation \(y = 3x + 1\):
   \[
   y = 3 \left(\frac{-3 + \sqrt{249}}{10}\right) + 1
   \]
   \[
   y = \frac{-9 + 3\sqrt{249}}{10} + 1
   \]
   \[
   y = \frac{-9 + 3\sqrt{249} + 10}{10}
   \]
   \[
   y = \frac{1 + 3\sqrt{249}}{10}
   \]

   Similarly,
   \[
   y = 3 \left(\frac{-3 - \sqrt{249}}{10}\right) + 1
   \]
   \[
   y = \frac{-9 - 3\sqrt{249}}{10} + 1
   \]
   \[
   y = \frac{-9 - 3\sqrt{249} + 10}{10}
   \]
   \[
   y = \frac{1 - 3\sqrt{249}}{10}
   \]

So, the points of intersection are:
\[
\left( \frac{-3 + \sqrt{249}}{10}, \frac{1 + 3\sqrt{249}}{10} \right) \quad \text{and} \quad \left( \frac{-3 - \sqrt{249}}{10}, \frac{1 - 3\sqrt{249}}{10} \right)
\]

### Summary:
- The circle's equation represents all points equidistant from a center point.
- The line's equation represents all points along a path with a constant slope.
- The points of intersection are found by solving the system of equations formed by the circle and the line.

Would you like more details or have any questions?

### Relative Questions:
1. How do you derive the general form of the equation of a circle from its center and radius?
2. What are the properties of the slope-intercept form of a linear equation?
3. Can a line intersect a circle at more than two points? Why or why not?
4. How does completing the square help in finding the center and radius of a circle given its general equation?
5. What are some real-life applications where solving the intersection of a circle and a line is useful?

### Tip:
When solving the intersection of a circle and a line, always check for extraneous solutions by substituting back into both original equations to verify the points.

Explore the difference between the equation of a circle and a line through a detailed example and solution. Learn how to find points of intersection between these geometric figures using algebraic methods. Suitable for advanced high school students and those interested in geometry and algebra.

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