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๐ Understanding the Equation of a Circle
The equation of a circle is a fundamental concept in geometry. It allows us to describe the circle's properties using algebra. Knowing how to find the equation of a circle, especially given the endpoints of its diameter, is a valuable skill. This guide provides a comprehensive, step-by-step explanation, complete with examples, to help you master this concept.
๐ A Brief History
The study of circles dates back to ancient civilizations, with mathematicians like Euclid exploring their properties extensively. The algebraic representation of geometric shapes, including the circle, became prominent with the development of analytic geometry by Renรฉ Descartes in the 17th century. This fusion of algebra and geometry paved the way for representing circles using equations.
๐ Key Principles
The standard form equation of a circle with center $(h, k)$ and radius $r$ is: $(x - h)^2 + (y - k)^2 = r^2$. When given the endpoints of a circle's diameter, we use these endpoints to find the center and the radius.
- ๐ Finding the Center: The center of the circle is the midpoint of the diameter. If the endpoints of the diameter are $(x_1, y_1)$ and $(x_2, y_2)$, the midpoint (center) is found using the midpoint formula: $h = \frac{x_1 + x_2}{2}$ and $k = \frac{y_1 + y_2}{2}$.
- Radius: The radius is half the length of the diameter. First, find the diameter length using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Then, the radius is $r = \frac{d}{2}$.
- ๐ Putting it Together: Once you have the center $(h, k)$ and the radius $r$, substitute these values into the standard form equation: $(x - h)^2 + (y - k)^2 = r^2$.
โ๏ธ Step-by-Step Guide
- ๐ Identify the Endpoints: Note down the coordinates of the diameter endpoints, $(x_1, y_1)$ and $(x_2, y_2)$.
- ๐งญ Calculate the Center: Use the midpoint formula to find the center $(h, k)$ of the circle: $h = \frac{x_1 + x_2}{2}$ and $k = \frac{y_1 + y_2}{2}$.
- ๐ Determine the Radius: Use the distance formula to find the length of the diameter and then divide by 2 to get the radius. $r = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2}$.
- โ๏ธ Write the Equation: Substitute the values of $h$, $k$, and $r$ into the standard equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$.
โ Real-World Examples
Example 1: Find the equation of a circle whose diameter endpoints are (2, 4) and (6, 8).
- Endpoints: $(x_1, y_1) = (2, 4)$ and $(x_2, y_2) = (6, 8)$.
- Center: $h = \frac{2 + 6}{2} = 4$ and $k = \frac{4 + 8}{2} = 6$. So, the center is (4, 6).
- Radius: $r = \frac{\sqrt{(6 - 2)^2 + (8 - 4)^2}}{2} = \frac{\sqrt{16 + 16}}{2} = \frac{\sqrt{32}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.
- Equation: $(x - 4)^2 + (y - 6)^2 = (2\sqrt{2})^2$, which simplifies to $(x - 4)^2 + (y - 6)^2 = 8$.
Example 2: Suppose a circular garden has markers at points (-1, 1) and (3, 5) representing the endpoints of its diameter. Determine the equation of the circle representing the garden.
- Endpoints: $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (3, 5)$.
- Center: $h = \frac{-1 + 3}{2} = 1$ and $k = \frac{1 + 5}{2} = 3$. So, the center is (1, 3).
- Radius: $r = \frac{\sqrt{(3 - (-1))^2 + (5 - 1)^2}}{2} = \frac{\sqrt{16 + 16}}{2} = \frac{\sqrt{32}}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.
- Equation: $(x - 1)^2 + (y - 3)^2 = (2\sqrt{2})^2$, which simplifies to $(x - 1)^2 + (y - 3)^2 = 8$.
๐ Practice Quiz
- โ Find the equation of the circle with diameter endpoints (1, 2) and (5, 6).
- ๐ The endpoints of a circle's diameter are (-2, 3) and (4, 1). What is the equation of the circle?
- ๐ Determine the equation of the circle given diameter endpoints (0, 0) and (4, 4).
- ๐ What is the equation of the circle if the diameter endpoints are (-1, -1) and (1, 1)?
- ๐ฏ A circle's diameter has endpoints (2, -3) and (2, 5). Find the equation of the circle.
๐ Conclusion
Finding the equation of a circle when given the endpoints of its diameter involves finding the circle's center (midpoint of the diameter) and radius (half the length of the diameter). These elements are then substituted into the standard equation of a circle. Understanding these steps enables the user to solve a wide range of geometry problems.
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