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.