Unit 10 Circles Homework 9 Standard Form Of A Circle

Let's delve into the fascinating world of circles, specifically focusing on how to express them in standard form and tackle homework problems related to this concept. The standard form of a circle provides a concise and powerful way to represent a circle's equation, making it easier to understand its properties and solve related problems.

Understanding the Standard Form of a Circle

The standard form of a circle's equation is:

(x - h)² + (y - k)² = r²

Where:

• (h, k) represents the coordinates of the circle's center.
• r represents the radius of the circle.

This equation is derived from the Pythagorean theorem. Imagine a right triangle formed by the radius of the circle (r), a horizontal line segment from the center to a point on the circle (x - h), and a vertical line segment from the center to the same point on the circle (y - k). Applying the Pythagorean theorem (a² + b² = c²) directly leads to the standard form equation.

Why Use the Standard Form?

The standard form offers several advantages:

• Easy Identification of Center and Radius: By simply looking at the equation, you can immediately identify the circle's center (h, k) and radius (r). This is crucial for graphing circles and solving related geometric problems.
• Simplified Problem Solving: Many circle-related problems become significantly easier to solve when the equation is in standard form.
• Direct Application of Geometric Principles: The standard form directly connects the equation to the circle's geometric properties, making it easier to apply geometric theorems and concepts.

Converting from General Form to Standard Form

Often, you'll encounter circle equations in the general form:

x² + y² + Dx + Ey + F = 0

To work effectively with these equations, you need to convert them to standard form. This is achieved through a process called completing the square.

Here's a step-by-step guide:

1. Rearrange the equation: Group the x terms together, the y terms together, and move the constant term to the right side of the equation:

   (x² + Dx) + (y² + Ey) = -F

2. Complete the square for the x terms: Take half of the coefficient of the x term (D/2), square it ((D/2)²), and add it to both sides of the equation.

   (x² + Dx + (D/2)²) + (y² + Ey) = -F + (D/2)²

3. Complete the square for the y terms: Take half of the coefficient of the y term (E/2), square it ((E/2)²), and add it to both sides of the equation.

   (x² + Dx + (D/2)²) + (y² + Ey + (E/2)²) = -F + (D/2)² + (E/2)²

4. Factor the perfect square trinomials: The expressions in parentheses are now perfect square trinomials, which can be factored as:

   (x + D/2)² + (y + E/2)² = -F + (D/2)² + (E/2)²

5. Simplify the right side: Simplify the constant term on the right side of the equation.

   (x + D/2)² + (y + E/2)² = (-4F + D² + E²) / 4

6. Identify the center and radius: Now the equation is in standard form. The center is (-D/2, -E/2), and the radius is √((-4F + D² + E²) / 4) or √(D² + E² - 4F) / 2.

Example:

Convert the following equation to standard form:

x² + y² - 6x + 4y - 3 = 0

1. Rearrange: (x² - 6x) + (y² + 4y) = 3
2. Complete the square for x: (x² - 6x + 9) + (y² + 4y) = 3 + 9
3. Complete the square for y: (x² - 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4
4. Factor: (x - 3)² + (y + 2)² = 16
5. Identify: Center: (3, -2), Radius: √16 = 4

Homework Problems and Solutions

Let's tackle some common homework problems related to the standard form of a circle.

Problem 1:

Write the equation of a circle with center (2, -3) and radius 5 in standard form.

Solution:

Using the standard form equation (x - h)² + (y - k)² = r², substitute h = 2, k = -3, and r = 5:

(x - 2)² + (y - (-3))² = 5²

(x - 2)² + (y + 3)² = 25

Problem 2:

Find the center and radius of the circle represented by the equation (x + 1)² + (y - 4)² = 9.

Solution:

Comparing the equation to the standard form, we can identify:

• h = -1 (since it's (x - h) and we have (x + 1))
• k = 4
• r² = 9, so r = √9 = 3

Therefore, the center is (-1, 4) and the radius is 3.

Problem 3:

Convert the following equation to standard form and find the center and radius:

x² + y² + 8x - 2y + 8 = 0

Solution:

1. Rearrange: (x² + 8x) + (y² - 2y) = -8
2. Complete the square for x: (x² + 8x + 16) + (y² - 2y) = -8 + 16
3. Complete the square for y: (x² + 8x + 16) + (y² - 2y + 1) = -8 + 16 + 1
4. Factor: (x + 4)² + (y - 1)² = 9
5. Identify: Center: (-4, 1), Radius: √9 = 3

Problem 4:

A circle has a center at (5, 2) and passes through the point (1, -1). Find the equation of the circle in standard form.

Solution:

We know the center (h, k) = (5, 2). We need to find the radius r. Since the circle passes through (1, -1), this point must satisfy the circle's equation. Substitute (x, y) = (1, -1) and (h, k) = (5, 2) into the standard form:

(1 - 5)² + (-1 - 2)² = r²

(-4)² + (-3)² = r²

16 + 9 = r²

25 = r²

Therefore, r = 5.

The equation of the circle in standard form is:

(x - 5)² + (y - 2)² = 25

Problem 5:

Determine whether the equation x² + y² - 4x + 6y + 15 = 0 represents a circle. If it does, find its center and radius.

Solution:

1. Rearrange: (x² - 4x) + (y² + 6y) = -15
2. Complete the square for x: (x² - 4x + 4) + (y² + 6y) = -15 + 4
3. Complete the square for y: (x² - 4x + 4) + (y² + 6y + 9) = -15 + 4 + 9
4. Factor: (x - 2)² + (y + 3)² = -2

Since the right side of the equation is negative (-2), it cannot be equal to r², which must be a non-negative value. Therefore, the equation does not represent a circle.

Problem 6:

The endpoints of a diameter of a circle are A(2, 5) and B(-4, -3). Find the equation of the circle in standard form.

Solution:

First, find the center of the circle, which is the midpoint of the diameter AB.

Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) = ((2 + (-4))/2, (5 + (-3))/2) = (-1, 1)

So, the center (h, k) = (-1, 1).

Next, find the radius of the circle. The radius is half the length of the diameter. We can find the length of the diameter using the distance formula between points A and B:

Distance = √((x₂ - x₁)² + (y₂ - y₁)²) = √((-4 - 2)² + (-3 - 5)²) = √((-6)² + (-8)²) = √(36 + 64) = √100 = 10

The diameter is 10, so the radius is r = 10/2 = 5.

The equation of the circle in standard form is:

(x + 1)² + (y - 1)² = 25

Problem 7:

Find the equation of the circle that is tangent to both the x-axis and the y-axis and has its center in the first quadrant and a radius of 3.

Solution:

Since the circle is tangent to both axes and its center is in the first quadrant, the coordinates of the center must be (r, r), where r is the radius. In this case, the radius is 3, so the center is (3, 3).

The equation of the circle in standard form is:

(x - 3)² + (y - 3)² = 9

Problem 8:

Given the equation (x - a)² + (y + b)² = c, describe how changing the values of a, b, and c affects the circle.

Solution:

• Changing 'a': Changing the value of 'a' shifts the circle horizontally. Increasing 'a' shifts the circle to the right, while decreasing 'a' shifts it to the left. The center of the circle changes from (a, -b) to a new x-coordinate.
• Changing 'b': Changing the value of 'b' shifts the circle vertically. Increasing 'b' shifts the circle down, while decreasing 'b' shifts it up. The center of the circle changes from (a, -b) to a new y-coordinate.
• Changing 'c': Changing the value of 'c' affects the radius of the circle. 'c' is equal to r², so increasing 'c' increases the radius (making the circle larger), and decreasing 'c' decreases the radius (making the circle smaller), as long as 'c' remains positive. If 'c' is negative, the equation does not represent a circle.

Problem 9:

Find the equation of the circle that passes through the points (0, 0), (2, 0), and (0, 2).

Solution:

Since the circle passes through (0, 0), (2, 0), and (0, 2), we can substitute these points into the general form of a circle equation: x² + y² + Dx + Ey + F = 0.

• (0, 0): 0² + 0² + D(0) + E(0) + F = 0 => F = 0
• (2, 0): 2² + 0² + D(2) + E(0) + 0 = 0 => 4 + 2D = 0 => D = -2
• (0, 2): 0² + 2² + D(0) + E(2) + 0 = 0 => 4 + 2E = 0 => E = -2

So, the general equation is x² + y² - 2x - 2y = 0.

Now, convert to standard form:

1. Rearrange: (x² - 2x) + (y² - 2y) = 0
2. Complete the square for x: (x² - 2x + 1) + (y² - 2y) = 0 + 1
3. Complete the square for y: (x² - 2x + 1) + (y² - 2y + 1) = 0 + 1 + 1
4. Factor: (x - 1)² + (y - 1)² = 2

The equation of the circle in standard form is (x - 1)² + (y - 1)² = 2.

Problem 10:

A circle has its center on the line y = x and is tangent to the x-axis at the point (4, 0). Find the equation of the circle in standard form.

Solution:

Since the circle is tangent to the x-axis at (4, 0), the x-coordinate of the center is 4, and the radius is equal to the y-coordinate of the center. Because the center lies on the line y = x, the y-coordinate of the center is also 4. Therefore, the center is (4, 4) and the radius is 4.

The equation of the circle in standard form is:

(x - 4)² + (y - 4)² = 16

Advanced Concepts and Applications

Beyond basic homework problems, the standard form of a circle is crucial in more advanced mathematical concepts, including:

• Analytic Geometry: Studying the intersections of circles with lines and other curves.
• Calculus: Finding tangents and normals to circles, and calculating areas and volumes related to circular regions.
• Complex Numbers: Representing circles in the complex plane.
• Computer Graphics: Drawing and manipulating circles in computer programs.

Common Mistakes to Avoid

• Incorrectly Identifying the Center: Remember that the center is (h, k) in the equation (x - h)² + (y - k)² = r². Pay attention to the signs! (x + 3)² means h = -3, not 3.
• Forgetting to Complete the Square Correctly: When converting from general form, double-check your work when completing the square. Adding the same value to both sides is crucial.
• Confusing Radius and Radius Squared: The equation gives you r², not r. Remember to take the square root to find the radius.
• Assuming All Equations Represent Circles: Equations that look like circles might not be. If, after completing the square, you get a negative value on the right side, the equation doesn't represent a circle.

Conclusion

Mastering the standard form of a circle and the techniques for converting equations into this form is essential for success in geometry and related mathematical fields. By understanding the underlying principles and practicing with various problems, you'll develop a strong foundation for tackling more complex challenges involving circles. Remember to pay attention to details, avoid common mistakes, and connect the equation to the geometric properties of the circle. With consistent effort, you'll find that working with circles becomes a rewarding and insightful experience.