Finding The Equation Of A Circle's Diameter With A Given Gradient
In this article, we will explore how to determine the equation of a diameter of a circle when given its gradient and the circle's equation. This problem combines concepts from coordinate geometry, including the standard equation of a circle and the properties of diameters. Understanding how to solve this type of problem is crucial for students studying analytical geometry and can be applied in various mathematical contexts.
Understanding the Circle's Equation
Before we dive into the problem, let's revisit the general equation of a circle. The general equation of a circle is given by:
x² + y² + 2gx + 2fy + c = 0
where (-g, -f) represents the center of the circle and the radius, r, is calculated as:
r = √(g² + f² - c)
To relate this to our specific problem, we need to rewrite the given equation in this standard form. This will allow us to easily identify the center of the circle, which is a crucial point for determining the equation of any diameter.
Transforming the Given Equation
Our given equation is:
x² + y² - x + 2y - 3 = 0
To transform this into the general form, we need to complete the square for both the x and y terms. This process involves rearranging the equation and adding and subtracting constants to create perfect square trinomials.
Let's start by grouping the x and y terms:
(x² - x) + (y² + 2y) = 3
Now, we complete the square for the x terms. To do this, we take half of the coefficient of the x term (-1), square it ((-1/2)² = 1/4), and add it to both sides:
(x² - x + 1/4) + (y² + 2y) = 3 + 1/4
Next, we complete the square for the y terms. We take half of the coefficient of the y term (2), square it ((2/2)² = 1), and add it to both sides:
(x² - x + 1/4) + (y² + 2y + 1) = 3 + 1/4 + 1
Now, we can rewrite the expressions in parentheses as perfect squares:
(x - 1/2)² + (y + 1)² = 3 + 1/4 + 1
Simplifying the right side:
(x - 1/2)² + (y + 1)² = 17/4
Identifying the Center and Radius
Now that we have the equation in the standard form:
(x - h)² + (y - k)² = r²
We can easily identify the center (h, k) and the radius r. Comparing our equation to the standard form, we see that:
h = 1/2 k = -1 r² = 17/4 r = √(17/4) = √17 / 2
Therefore, the center of the circle is (1/2, -1) and the radius is √17 / 2. This information is crucial for finding the equation of the diameter.
Finding the Equation of the Diameter
The problem states that one diameter of the circle has a gradient (slope) of 1/4. A diameter of a circle is a line segment that passes through the center of the circle and has endpoints on the circle's circumference. Thus, any diameter must pass through the center of the circle.
Using the Point-Slope Form
Since we know the gradient (m = 1/4) and a point that the diameter passes through (the center of the circle (1/2, -1)), we can use the point-slope form of a linear equation to find the equation of the diameter. The point-slope form is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) is a point on the line and m is the slope.
Substituting the center (1/2, -1) and the gradient 1/4 into the point-slope form, we get:
y - (-1) = (1/4)(x - 1/2)
Simplifying:
y + 1 = (1/4)(x - 1/2)
Converting to Slope-Intercept Form
To make the equation more readable and easier to work with, we can convert it to the slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Multiplying both sides of the equation by 4 to eliminate the fraction:
4(y + 1) = x - 1/2
Distributing the 4 on the left side:
4y + 4 = x - 1/2
Subtracting 4 from both sides:
4y = x - 1/2 - 4
Simplifying the right side:
4y = x - 9/2
Dividing both sides by 4:
y = (1/4)x - 9/8
This is the equation of the diameter in slope-intercept form.
General Form of the Equation
We can also express the equation in the general form (Ax + By + C = 0). To do this, we start with the slope-intercept form:
y = (1/4)x - 9/8
Multiply both sides by 8 to eliminate fractions:
8y = 2x - 9
Rearrange the terms to get the general form:
2x - 8y - 9 = 0
Thus, the equation of the diameter can be expressed in general form as 2x - 8y - 9 = 0.
Conclusion
In summary, we have successfully found the equation of the diameter of the circle x² + y² - x + 2y - 3 = 0 with a gradient of 1/4. We began by transforming the given equation into the standard form of a circle's equation, which allowed us to identify the center of the circle. Knowing the center and the gradient, we used the point-slope form to find the equation of the diameter and then converted it into both slope-intercept and general forms. This problem demonstrates the importance of understanding the properties of circles and linear equations in coordinate geometry. Mastering these concepts is essential for solving a wide range of mathematical problems.
Key takeaways:
- The general equation of a circle is x² + y² + 2gx + 2fy + c = 0, and the standard form is (x - h)² + (y - k)² = r².
- The center of the circle can be found by completing the square and rewriting the equation in standard form.
- A diameter of a circle passes through the center.
- The point-slope form of a linear equation is y - y₁ = m(x - x₁).
- The slope-intercept form of a linear equation is y = mx + b.
- The general form of a linear equation is Ax + By + C = 0.
By following these steps and understanding the underlying concepts, you can confidently tackle similar problems involving circles and their diameters. Remember to practice and apply these techniques to reinforce your understanding and improve your problem-solving skills.
