Lesson 1: Circles
Learning Objectives
- Understand conic sections and how circles are formed by slicing a cone
- Define a circle as the set of all points equidistant from a center point
- Write and recognize the standard form equation of a circle
- Find the center and radius of a circle from its equation
- Write the equation of a circle given its center and radius
- Graph circles by plotting the center and using the radius
- Convert from general form to standard form using completing the square
- Find x-intercepts and y-intercepts of circles
- Apply circle equations to real-world problems involving distance and design
Introduction: What are Conic Sections?
Conic Sections: Curves formed by the intersection of a plane and a double-napped cone (two cones placed tip-to-tip).
Depending on the angle at which the plane intersects the cone, we get different curves:
- Circle: Plane cuts perpendicular to the axis of the cone
- Ellipse: Plane cuts at an angle, but does not pass through the base
- Parabola: Plane cuts parallel to the side of the cone
- Hyperbola: Plane cuts through both cones
In this lesson, we focus on circles, the simplest and most symmetric of all conic sections.
Circle (Geometric Definition): The set of all points in a plane that are equidistant from a fixed point called the center. The constant distance is called the radius.
Section 1: Standard Form of a Circle
To derive the equation of a circle, we use the distance formula. If a point (x, y) is on a circle with center (h, k) and radius r, then the distance from (x, y) to (h, k) equals r.
Standard Form of a Circle:
(x - h)² + (y - k)² = r²
where (h, k) is the center and r is the radius
Special Case: A circle centered at the origin (0, 0) has the equation:
x² + y² = r²
Example 1: Identifying Center and Radius from Standard Form
Find the center and radius of the circle: (x - 3)² + (y + 2)² = 25
Solution:
Compare to standard form (x - h)² + (y - k)² = r²:
(x - 3)² means h = 3
(y + 2)² = (y - (-2))² means k = -2
r² = 25, so r = 5
Answer: Center: (3, -2), Radius: 5
Example 2: Circle Centered at the Origin
Find the center and radius of the circle: x² + y² = 49
Solution:
This is the special form x² + y² = r²:
Center: (0, 0) (the origin)
r² = 49, so r = 7
Answer: Center: (0, 0), Radius: 7
Example 3: Careful with Signs
Find the center and radius of the circle: (x + 5)² + (y - 1)² = 16
Solution:
Rewrite in standard form to identify h and k:
(x + 5)² = (x - (-5))², so h = -5
(y - 1)² means k = 1
r² = 16, so r = 4
Answer: Center: (-5, 1), Radius: 4
Common Mistake
Be careful with signs! In the equation (x - h)² + (y - k)² = r², the center coordinates have the OPPOSITE sign from what appears in the equation:
- (x - 3)² means h = +3
- (x + 3)² means h = -3
Example 4: Writing the Equation Given Center and Radius
Write the equation of a circle with center (2, -4) and radius 6.
Solution:
Use the standard form (x - h)² + (y - k)² = r²:
h = 2, k = -4, r = 6
Substitute:
(x - 2)² + (y - (-4))² = 6²
(x - 2)² + (y + 4)² = 36
Answer: (x - 2)² + (y + 4)² = 36
Example 5: Circle with Center at Origin
Write the equation of a circle centered at the origin with radius 10.
Solution:
For a circle at the origin, use x² + y² = r²:
x² + y² = 10²
x² + y² = 100
Answer: x² + y² = 100
Example 6: Using Fractional Radius
Find the center and radius of: (x - 1/2)² + (y + 3/4)² = 9/16
Solution:
Identify h, k, and r:
h = 1/2
k = -3/4
r² = 9/16, so r = 3/4
Answer: Center: (1/2, -3/4), Radius: 3/4
Section 2: Graphing Circles
To graph a circle, we need only two pieces of information: the center and the radius.
Steps for Graphing a Circle:
- Identify the center (h, k) and plot it
- From the center, move r units in each direction (up, down, left, right)
- Plot these four points
- Sketch a smooth circle through these points
Example 7: Graphing a Circle
Graph the circle: (x - 1)² + (y - 2)² = 9
Solution:
Step 1: Identify center and radius.
Center: (1, 2)
r² = 9, so r = 3
Step 2: Plot the center at (1, 2).
Step 3: From (1, 2), move 3 units in each direction:
Right: (1 + 3, 2) = (4, 2)
Left: (1 - 3, 2) = (-2, 2)
Up: (1, 2 + 3) = (1, 5)
Down: (1, 2 - 3) = (1, -1)
Step 4: Draw a smooth circle through these four points.
Answer: Circle centered at (1, 2) with radius 3, passing through (4, 2), (-2, 2), (1, 5), and (1, -1).
Example 8: Graphing with Negative Coordinates
Graph the circle: (x + 3)² + (y + 1)² = 16
Solution:
Step 1: Identify center and radius.
Center: (-3, -1)
r = 4
Step 2: From (-3, -1), move 4 units in each direction:
Right: (-3 + 4, -1) = (1, -1)
Left: (-3 - 4, -1) = (-7, -1)
Up: (-3, -1 + 4) = (-3, 3)
Down: (-3, -1 - 4) = (-3, -5)
Answer: Circle centered at (-3, -1) with radius 4.
Example 9: Writing Equation from Graph
A circle is graphed with center at (-2, 5) and passes through the point (1, 5). Write its equation.
Solution:
Step 1: The center is given: h = -2, k = 5.
Step 2: Find the radius using the distance from center to the given point:
r = distance from (-2, 5) to (1, 5)
r = √[(1 - (-2))² + (5 - 5)²]
r = √[3² + 0²]
r = √9 = 3
Step 3: Write the equation:
(x - (-2))² + (y - 5)² = 3²
(x + 2)² + (y - 5)² = 9
Answer: (x + 2)² + (y - 5)² = 9
Example 10: Finding Radius from Two Points
Write the equation of a circle with center at (0, 0) that passes through the point (3, 4).
Solution:
Step 1: Center is at origin: h = 0, k = 0.
Step 2: Find radius using distance formula:
r = √[(3 - 0)² + (4 - 0)²]
r = √[9 + 16]
r = √25 = 5
Step 3: Write the equation:
x² + y² = 5²
x² + y² = 25
Answer: x² + y² = 25
Section 3: General Form and Completing the Square
Sometimes circles are given in general form rather than standard form:
General Form of a Circle:
x² + y² + Dx + Ey + F = 0
To find the center and radius, we must convert to standard form by completing the square.
Steps for Completing the Square:
- Group x terms together and y terms together
- Move the constant to the right side
- Complete the square for x: add (D/2)² to both sides
- Complete the square for y: add (E/2)² to both sides
- Factor the perfect squares and simplify
Example 11: Converting General Form to Standard Form
Find the center and radius: x² + y² + 6x - 8y + 9 = 0
Solution:
Step 1: Group x and y terms:
(x² + 6x) + (y² - 8y) = -9
Step 2: Complete the square for x. Take half of 6, square it: (6/2)² = 9
(x² + 6x + 9) + (y² - 8y) = -9 + 9
Step 3: Complete the square for y. Take half of -8, square it: (-8/2)² = 16
(x² + 6x + 9) + (y² - 8y + 16) = -9 + 9 + 16
Step 4: Factor and simplify:
(x + 3)² + (y - 4)² = 16
Answer: Center: (-3, 4), Radius: 4
Example 12: Completing the Square with Negative Coefficients
Convert to standard form: x² + y² - 4x + 10y + 13 = 0
Solution:
Step 1: Rearrange:
(x² - 4x) + (y² + 10y) = -13
Step 2: Complete the square for x: (-4/2)² = 4
(x² - 4x + 4) + (y² + 10y) = -13 + 4
Step 3: Complete the square for y: (10/2)² = 25
(x² - 4x + 4) + (y² + 10y + 25) = -13 + 4 + 25
Step 4: Factor:
(x - 2)² + (y + 5)² = 16
Answer: Center: (2, -5), Radius: 4
Example 13: When the Result is Not a Circle
Determine whether x² + y² + 2x - 4y + 10 = 0 represents a circle.
Solution:
Complete the square:
(x² + 2x) + (y² - 4y) = -10
(x² + 2x + 1) + (y² - 4y + 4) = -10 + 1 + 4
(x + 1)² + (y - 2)² = -5
Since r² = -5, which is negative, this equation has no real solutions.
Answer: This does NOT represent a circle (the radius cannot be imaginary).
Example 14: Circle Reduces to a Point
Find the center and radius: x² + y² - 6x + 2y + 10 = 0
Solution:
Complete the square:
(x² - 6x) + (y² + 2y) = -10
(x² - 6x + 9) + (y² + 2y + 1) = -10 + 9 + 1
(x - 3)² + (y + 1)² = 0
This means r = 0.
Answer: This is a degenerate circle (a single point) at (3, -1).
Example 15: Complete Process Example
Find the center and radius, then graph: x² + y² + 8x - 6y = 0
Solution:
Step 1: Complete the square:
(x² + 8x) + (y² - 6y) = 0
(x² + 8x + 16) + (y² - 6y + 9) = 16 + 9
(x + 4)² + (y - 3)² = 25
Step 2: Identify center and radius:
Center: (-4, 3)
Radius: 5
Answer: Center: (-4, 3), Radius: 5. Graph by plotting center and marking points 5 units away in all directions.
Section 4: Finding Intercepts
Intercepts are points where the circle crosses the x-axis or y-axis.
Finding Intercepts:
- x-intercepts: Set y = 0 and solve for x
- y-intercepts: Set x = 0 and solve for y
Example 16: Finding x-intercepts
Find the x-intercepts of the circle: x² + y² - 4x - 2y - 4 = 0
Solution:
Step 1: Set y = 0:
x² + 0² - 4x - 2(0) - 4 = 0
x² - 4x - 4 = 0
Step 2: Use the quadratic formula:
x = [4 ± √(16 + 16)] / 2
x = [4 ± √32] / 2
x = [4 ± 4√2] / 2
x = 2 ± 2√2
Answer: x-intercepts: (2 + 2√2, 0) and (2 - 2√2, 0), or approximately (4.83, 0) and (-0.83, 0)
Example 17: Finding y-intercepts
Find the y-intercepts of the circle: x² + y² = 25
Solution:
Step 1: Set x = 0:
0² + y² = 25
y² = 25
y = ±5
Answer: y-intercepts: (0, 5) and (0, -5)
Example 18: Circle with No x-intercepts
Determine if the circle (x - 1)² + (y - 5)² = 9 has x-intercepts.
Solution:
Step 1: Set y = 0:
(x - 1)² + (0 - 5)² = 9
(x - 1)² + 25 = 9
(x - 1)² = -16
Since (x - 1)² cannot be negative, there are no real solutions.
Answer: This circle has NO x-intercepts. (The circle is centered at (1, 5) with radius 3, so it's entirely above the x-axis.)
Example 19: Finding All Intercepts
Find all intercepts: x² + y² - 6x + 4y = 0
Solution:
For x-intercepts (set y = 0):
x² - 6x = 0
x(x - 6) = 0
x = 0 or x = 6
x-intercepts: (0, 0) and (6, 0)
For y-intercepts (set x = 0):
y² + 4y = 0
y(y + 4) = 0
y = 0 or y = -4
y-intercepts: (0, 0) and (0, -4)
Answer: x-intercepts: (0, 0) and (6, 0); y-intercepts: (0, 0) and (0, -4). Note: (0, 0) is both an x-intercept and y-intercept.
Section 5: Applications of Circles
Circles appear in many real-world contexts, from engineering and design to navigation and communication.
Example 20: Wireless Signal Range
A cell phone tower is located at coordinates (2, 3) on a city map (in miles). The tower has a signal range of 5 miles. Write an equation representing the boundary of the signal coverage.
Solution:
The signal coverage forms a circle with:
Center: (2, 3)
Radius: 5 miles
Write the equation:
(x - 2)² + (y - 3)² = 25
Answer: (x - 2)² + (y - 3)² = 25
Example 21: Circular Garden Design
A landscape designer wants to create a circular flower garden centered at the point (4, -1) in a coordinate system where units are in feet. The garden should have a diameter of 12 feet. Write the equation of the garden's boundary.
Solution:
Step 1: Identify center and radius:
Center: (4, -1)
Diameter = 12 feet, so radius = 6 feet
Step 2: Write the equation:
(x - 4)² + (y + 1)² = 36
Answer: (x - 4)² + (y + 1)² = 36
Example 22: Distance and Circles
An earthquake epicenter is located at (-3, 2) on a regional map. Seismographs detect the earthquake at a distance of 10 km from the epicenter. Is a city located at (5, 8) affected by the earthquake?
Solution:
Step 1: Find the distance from epicenter (-3, 2) to city (5, 8):
d = √[(5 - (-3))² + (8 - 2)²]
d = √[8² + 6²]
d = √[64 + 36]
d = √100 = 10 km
Step 2: Compare to the earthquake range of 10 km.
Since the distance equals exactly 10 km, the city is on the boundary of the affected area.
Answer: Yes, the city is affected (it's on the edge of the 10 km radius).
Example 23: Circular Running Track
A circular running track has a center at the origin and passes through the point (50, 0), where distances are in meters. Find the equation of the track and determine if a water fountain at (30, 40) is inside, outside, or on the track.
Solution:
Step 1: Find the radius (distance from origin to (50, 0)):
r = √[(50 - 0)² + (0 - 0)²] = 50 meters
Step 2: Write the equation:
x² + y² = 2500
Step 3: Check if (30, 40) satisfies the equation:
30² + 40² = 900 + 1600 = 2500
Since the equation is satisfied, the point is ON the track.
Answer: Equation: x² + y² = 2500. The fountain at (30, 40) is exactly on the track.
Example 24: Sprinkler Coverage
A sprinkler system covers a circular area described by the equation x² + y² - 10x + 6y - 2 = 0, where x and y are in feet. Find the center of the sprinkler and its coverage radius.
Solution:
Complete the square to find center and radius:
(x² - 10x) + (y² + 6y) = 2
(x² - 10x + 25) + (y² + 6y + 9) = 2 + 25 + 9
(x - 5)² + (y + 3)² = 36
Answer: The sprinkler is centered at (5, -3) with a coverage radius of 6 feet.
Example 25: Circular Pathway
A park has a circular walking path. Two benches are located at points A(1, 7) and B(9, 7), which are endpoints of a diameter of the circle. Find the equation of the circular path.
Solution:
Step 1: Find the center (midpoint of diameter):
h = (1 + 9)/2 = 5
k = (7 + 7)/2 = 7
Center: (5, 7)
Step 2: Find the radius (distance from center to either endpoint):
r = √[(9 - 5)² + (7 - 7)²]
r = √[16 + 0] = 4
Step 3: Write the equation:
(x - 5)² + (y - 7)² = 16
Answer: (x - 5)² + (y - 7)² = 16
Check Your Understanding
1. Find the center and radius of the circle: (x - 7)² + (y + 3)² = 64
Answer: Center: (7, -3), Radius: 8
From standard form: h = 7, k = -3, r² = 64 so r = 8
2. Write the equation of a circle with center (-2, 5) and radius 3.
Answer: (x + 2)² + (y - 5)² = 9
Use (x - h)² + (y - k)² = r² with h = -2, k = 5, r = 3
3. Convert to standard form: x² + y² - 8x + 12y + 3 = 0
Answer: (x - 4)² + (y + 6)² = 49
Complete the square: (x² - 8x + 16) + (y² + 12y + 36) = -3 + 16 + 36 = 49
4. Find the center and radius: x² + y² = 81
Answer: Center: (0, 0), Radius: 9
This is a circle centered at the origin with r² = 81, so r = 9
5. Does the equation x² + y² + 4x - 6y + 20 = 0 represent a circle?
Answer: No
Completing the square: (x + 2)² + (y - 3)² = -7. Since r² is negative, this is not a real circle.
6. Find the y-intercepts of x² + y² - 6x = 0
Answer: (0, 0) only
Set x = 0: y² = 0, so y = 0. The only y-intercept is the origin.
7. A circle has center at (3, -2) and passes through (6, 2). Find its equation.
Answer: (x - 3)² + (y + 2)² = 25
r = √[(6-3)² + (2-(-2))²] = √[9 + 16] = 5, so r² = 25
8. What are the x-intercepts of (x - 2)² + (y - 1)² = 4?
Answer: (2 + √3, 0) and (2 - √3, 0)
Set y = 0: (x - 2)² + 1 = 4, so (x - 2)² = 3, giving x = 2 ± √3
9. Find the center and radius: x² + y² + 10y = 0
Answer: Center: (0, -5), Radius: 5
x² + (y² + 10y + 25) = 25, so x² + (y + 5)² = 25
10. Two points on a circle are (0, 4) and (6, 4), which are endpoints of a diameter. Find the equation.
Answer: (x - 3)² + (y - 4)² = 9
Center = midpoint = (3, 4); radius = half the diameter = 3
Key Takeaways
- A circle is the set of all points equidistant from a center point; that distance is the radius
- Standard form of a circle: (x - h)² + (y - k)² = r² where (h, k) is the center and r is the radius
- For a circle at the origin: x² + y² = r²
- Be careful with signs: (x - 3)² has center x-coordinate h = 3, while (x + 3)² has h = -3
- To graph a circle, plot the center and move r units in all four directions (up, down, left, right)
- General form is x² + y² + Dx + Ey + F = 0; convert to standard form by completing the square
- When completing the square, add (coefficient/2)² to both sides for each variable
- If r² is negative after completing the square, the equation does not represent a real circle
- If r² = 0, the circle is a single point (degenerate circle)
- Find intercepts by setting the other variable to zero and solving
- Circles model many real-world phenomena: signal range, coverage areas, circular paths, distances