Unit 10 Circles Homework 8 Equations Of Circles Answer Key

Circles, with their elegant symmetry, are a fundamental concept in geometry. Understanding their equations is crucial for success in mathematics and various applications. This article provides a comprehensive guide to understanding and solving equations of circles, particularly focusing on the types of problems you might encounter in Unit 10, Homework 8, related to equations of circles. We'll break down the concepts, provide step-by-step solutions, and offer tips for mastering this important topic.

Understanding the Equation of a Circle

The equation of a circle is derived from the Pythagorean theorem and represents all the points that are a fixed distance (the radius) from a central point. The standard form of the equation is:

(x - h)² + (y - k)² = r²

Where:

• (x, y) represents any point on the circle.
• (h, k) represents the coordinates of the center of the circle.
• r represents the radius of the circle.

This equation tells us that for any point (x, y) on the circle, the square of the horizontal distance from the center (x - h) plus the square of the vertical distance from the center (y - k) is equal to the square of the radius.

Key Concepts and Components

Before diving into specific problem types, let's solidify our understanding of the key components:

• Center (h, k): The center is the anchor of the circle. Its coordinates directly influence the equation. Remember that the values of h and k appear with opposite signs in the equation. For example, if the center is at (2, -3), the equation will include (x - 2) and (y + 3).

• Radius (r): The radius is the distance from the center to any point on the circle. It determines the size of the circle. The equation uses r², so if you're given the area of the circle, you'll need to find the radius by taking the square root of (Area / π). If you are given the diameter, divide by 2 to get the radius.

• General Form: While the standard form is most useful for identifying the center and radius, the general form is sometimes encountered:

  x² + y² + Dx + Ey + F = 0

  To work with the general form, you'll need to complete the square to convert it back to the standard form. This involves manipulating the equation to create perfect square trinomials for both the x and y terms.

Solving Common Problems in Unit 10 Homework 8

Unit 10 Homework 8 is likely to cover a variety of problem types relating to the equation of circles. Here are some examples and strategies for solving them:

1. Writing the Equation of a Circle Given the Center and Radius:

This is the most basic type of problem. You are given the coordinates of the center (h, k) and the radius (r), and your task is to plug these values into the standard form equation.

Example:

• Center: (3, -2)
• Radius: 5

Solution:

Using the standard form equation: (x - h)² + (y - k)² = r²

Substitute the given values: (x - 3)² + (y - (-2))² = 5²

Simplify: (x - 3)² + (y + 2)² = 25

2. Finding the Center and Radius Given the Equation of a Circle:

This is the reverse of the previous problem. You are given the equation in standard form, and you need to identify the center and radius.

Example:

Equation: (x + 1)² + (y - 4)² = 9

Solution:

Comparing to the standard form (x - h)² + (y - k)² = r², we can identify:

• h = -1 (Remember the sign changes: x + 1 means x - (-1))
• k = 4
• r² = 9, so r = √9 = 3

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

3. Writing the Equation of a Circle Given the Center and a Point on the Circle:

In this scenario, you are given the center (h, k) and a point (x, y) that lies on the circle. You'll need to use this information to find the radius first.

Steps:

1. Use the distance formula (derived from the Pythagorean theorem) to find the radius:

   r = √((x - h)² + (y - k)²)

   Where (x, y) is the point on the circle and (h, k) is the center.

2. Plug the center (h, k) and the calculated radius (r) into the standard form equation:

   (x - h)² + (y - k)² = r²

Example:

• Center: (2, 1)
• Point on the circle: (5, 5)

Solution:

1. Calculate the radius:

   r = √((5 - 2)² + (5 - 1)²) = √(3² + 4²) = √(9 + 16) = √25 = 5

2. Write the equation:

   (x - 2)² + (y - 1)² = 5²

   (x - 2)² + (y - 1)² = 25

4. Writing the Equation of a Circle Given the Endpoints of a Diameter:

When given the endpoints of a diameter, you'll need to find the center (the midpoint of the diameter) and the radius (half the length of the diameter).

Steps:

1. Find the center (h, k) using the midpoint formula:

   h = (x₁ + x₂) / 2 k = (y₁ + y₂) / 2

   Where (x₁, y₁) and (x₂, y₂) are the endpoints of the diameter.

2. Find the radius (r) using the distance formula between the center and one of the endpoints:

   r = √((x - h)² + (y - k)²)

   Where (x, y) is one of the endpoints and (h, k) is the center. Alternatively, find the full length of the diameter using the distance formula between the endpoints and divide by 2.

3. Plug the center (h, k) and the radius (r) into the standard form equation:

   (x - h)² + (y - k)² = r²

Example:

• Endpoints of diameter: (1, 2) and (5, 6)

Solution:

1. Calculate the center:

   h = (1 + 5) / 2 = 3 k = (2 + 6) / 2 = 4

   Center: (3, 4)

2. Calculate the radius (using endpoint (1,2) and center (3,4)):

   r = √((1 - 3)² + (2 - 4)²) = √((-2)² + (-2)²) = √(4 + 4) = √8 = 2√2

3. Write the equation:

   (x - 3)² + (y - 4)² = (2√2)²

   (x - 3)² + (y - 4)² = 8

5. Converting from General Form to Standard Form:

This is a more advanced problem that requires completing the square.

Steps:

1. Group the x terms and 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. Rewrite the x terms and y terms as squared binomials:

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

5. Simplify the right side of the equation to find r²:

   (x + D/2)² + (y + E/2)² = r²

   Now the equation is in standard form, and you can identify the center and radius. Remember that the center coordinates will be (-D/2, -E/2).

Example:

Equation: x² + y² + 4x - 6y - 12 = 0

Solution:

1. Group terms:

   x² + 4x + y² - 6y = 12

2. Complete the square for x: (4/2)² = 2² = 4

   x² + 4x + 4 + y² - 6y = 12 + 4

3. Complete the square for y: (-6/2)² = (-3)² = 9

   x² + 4x + 4 + y² - 6y + 9 = 12 + 4 + 9

4. Rewrite as squared binomials:

   (x + 2)² + (y - 3)² = 25

5. Identify center and radius:

   Center: (-2, 3) Radius: √25 = 5

6. Determining if a Point Lies on, Inside, or Outside the Circle:

Given the equation of a circle and a point (x, y), you can determine its location relative to the circle.

Steps:

1. Plug the coordinates of the point (x, y) into the left side of the circle's equation:

   (x - h)² + (y - k)²

2. Compare the result to r²:

   • If (x - h)² + (y - k)² = r², the point lies on the circle.
   • If (x - h)² + (y - k)² < r², the point lies inside the circle.
   • If (x - h)² + (y - k)² > r², the point lies outside the circle.

Example:

• Equation: (x - 1)² + (y + 2)² = 16
• Point: (3, -1)

Solution:

1. Plug in the point:

   (3 - 1)² + (-1 + 2)² = 2² + 1² = 4 + 1 = 5

2. Compare to r² (which is 16):

   5 < 16

Therefore, the point (3, -1) lies inside the circle.

Tips for Success

• Master the standard form equation: Know it inside and out. Understand how changing h, k, and r affects the circle's position and size.
• Practice completing the square: This is a crucial skill for dealing with equations in general form.
• Visualize the circle: Sketch a quick graph of the circle and the given information (center, point, etc.). This can help you understand the problem and avoid mistakes.
• Double-check your work: Carefully review your calculations, especially when dealing with signs.
• Understand the relationship between the equation and the geometry: Remember that the equation is just a mathematical representation of a geometric shape. Connecting the two will deepen your understanding.
• Practice, practice, practice: The more problems you solve, the more comfortable you'll become with the concepts and techniques.

Common Mistakes to Avoid

• Forgetting to square the radius: The equation uses r², not r.
• Incorrect signs for the center: Remember that the center coordinates appear with opposite signs in the equation (x - h) and (y - k).
• Errors in completing the square: Be careful with the arithmetic when completing the square, especially when dealing with fractions.
• Misinterpreting the general form: Don't try to directly identify the center and radius from the general form. Convert it to standard form first.
• Using the diameter instead of the radius: Make sure you're using the radius in the equation, not the diameter.

Example Problems and Detailed Solutions

Let's work through a few more detailed examples to solidify your understanding.

Problem 1:

Write the equation of a circle that is tangent to the x-axis, has its center on the line y = x, and a radius of 4.

Solution:

1. Understanding Tangency: If a circle is tangent to the x-axis, it means the circle touches the x-axis at exactly one point. This implies that the absolute value of the y-coordinate of the center is equal to the radius.

2. Center on the Line y = x: Since the center lies on the line y = x, the x-coordinate and y-coordinate of the center are equal. Let the center be (a, a).

3. Using the Radius: We know the radius is 4, and since the circle is tangent to the x-axis, |a| = 4. This gives us two possibilities for the center: (4, 4) or (-4, -4). However, a circle centered at (-4, -4) with a radius of 4 would not be tangent to the positive x-axis (it would be tangent to the negative x-axis). Therefore, the center must be (4, 4).

4. Writing the Equation: Using the standard form equation with center (4, 4) and radius 4:

   (x - 4)² + (y - 4)² = 4² (x - 4)² + (y - 4)² = 16

Problem 2:

Find the equation of the circle passing through the points (1, 1), (5, 1), and (4, 4).

Solution:

This problem requires a more strategic approach. We can't directly use the standard form because we don't know the center or radius. We'll use the fact that the distance from the center to each of these points is the same (the radius).

1. General Equation: Start with the general form of the circle's equation:

   x² + y² + Dx + Ey + F = 0

2. Substitute the Points: Substitute each point into the equation to create a system of three equations:

   • For (1, 1): 1² + 1² + D(1) + E(1) + F = 0 => D + E + F = -2
   • For (5, 1): 5² + 1² + D(5) + E(1) + F = 0 => 5D + E + F = -26
   • For (4, 4): 4² + 4² + D(4) + E(4) + F = 0 => 4D + 4E + F = -32

3. Solve the System of Equations: Solve the system of equations for D, E, and F. This can be done using substitution, elimination, or matrices. Here's one way to solve using elimination:

   • Subtract the first equation from the second: (5D + E + F) - (D + E + F) = -26 - (-2) => 4D = -24 => D = -6
   • Subtract the first equation from the third: (4D + 4E + F) - (D + E + F) = -32 - (-2) => 3D + 3E = -30 => D + E = -10
   • Substitute D = -6 into D + E = -10: -6 + E = -10 => E = -4
   • Substitute D = -6 and E = -4 into D + E + F = -2: -6 - 4 + F = -2 => F = 8

4. Write the General Equation: Substitute the values of D, E, and F back into the general equation:

   x² + y² - 6x - 4y + 8 = 0

5. Convert to Standard Form: Complete the square to convert the general form to standard form:

   • (x² - 6x) + (y² - 4y) = -8
   • (x² - 6x + 9) + (y² - 4y + 4) = -8 + 9 + 4
   • (x - 3)² + (y - 2)² = 5

Therefore, the equation of the circle is (x - 3)² + (y - 2)² = 5. The center is (3, 2) and the radius is √5.

Problem 3:

Find the equation of the line tangent to the circle (x - 2)² + (y + 1)² = 25 at the point (5, 3).

Solution:

This problem combines our knowledge of circles with linear equations.

1. Find the Slope of the Radius: The tangent line is perpendicular to the radius at the point of tangency. First, find the slope of the radius connecting the center (2, -1) and the point (5, 3):

   m_radius = (3 - (-1)) / (5 - 2) = 4 / 3

2. Find the Slope of the Tangent Line: Since the tangent line is perpendicular to the radius, its slope is the negative reciprocal of the radius's slope:

   m_tangent = -3 / 4

3. Write the Equation of the Tangent Line: Use the point-slope form of a line (y - y₁ = m(x - x₁)) with the point (5, 3) and the slope -3/4:

   y - 3 = (-3/4)(x - 5)

4. Simplify the Equation: Convert the equation to slope-intercept form (y = mx + b) or standard form (Ax + By = C):

   y - 3 = (-3/4)x + 15/4 y = (-3/4)x + 15/4 + 3 y = (-3/4)x + 15/4 + 12/4 y = (-3/4)x + 27/4

Therefore, the equation of the tangent line is y = (-3/4)x + 27/4. We could also write this as 3x + 4y = 27.

Conclusion

Mastering the equations of circles requires a solid understanding of the standard form, the ability to manipulate equations through completing the square, and the application of related geometric concepts. By working through practice problems, understanding the underlying principles, and avoiding common mistakes, you can confidently tackle Unit 10 Homework 8 and any other challenges involving circles. Remember to visualize the problems, double-check your work, and connect the equations to the geometric representations to deepen your understanding and improve your problem-solving skills. Good luck!