6.2 Equation of a Circle
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Learning Objectives
- Understand the standard form of a circle's equation
- Sketch circles from equations and vice versa
- Find the centre and radius from a circle equation
- Solve geometric problems involving circles
- Apply circle equations to real-world scenarios
Key Concepts
Standard Equation of a Circle
- A circle with **centre (a, b)** and **radius r** has the equation:
(x - a)² + (y - b)² = r²
- If the circle is centred at the origin (0, 0), the equation is:
x² + y² = r²
Finding Centre and Radius
- Rearranged equations can be used to identify the centre and radius.
- Example: (x - 3)² + (y + 2)² = 25 →
- Centre: (3, -2)
- Radius: √25 = 5
Example 1
Find the centre and radius of the circle (x - 4)² + (y + 1)² = 36
Solution:
Step 1: Compare with standard form (x - a)² + (y - b)² = r² Step 2: Identify the centre: - x-coordinate: a = 4 (from x - 4) - y-coordinate: b = -1 (from y + 1 = y - (-1)) - Centre: (4, -1) Step 3: Find the radius: - r² = 36 - r = √36 = 6 So the circle has centre (4, -1) and radius 6.
Example 2
Find the centre and radius of the circle x² + y² = 64
Solution:
Step 1: This is in the form x² + y² = r² (centre at origin) Step 2: Identify the centre: - Since there are no x or y terms, a = 0 and b = 0 - Centre: (0, 0) Step 3: Find the radius: - r² = 64 - r = √64 = 8 So the circle has centre (0, 0) and radius 8.
Expanding and Completing the Square
- You might need to complete the square to rewrite the circle equation into standard form.
- Example: x² + y² + 4x - 6y + 9 = 0 →
Group terms: (x² + 4x) + (y² - 6y) = -9
Complete the square:
(x + 2)² - 4 + (y - 3)² - 9 = -9
→ (x + 2)² + (y - 3)² = 4
- Centre: (-2, 3), Radius: √4 = 2
Example 1
Convert x² + y² + 6x - 8y + 9 = 0 to standard form and find the centre and radius
Solution:
Step 1: Group x and y terms: (x² + 6x) + (y² - 8y) = -9 Step 2: Complete the square for x: x² + 6x = (x + 3)² - 9 Step 3: Complete the square for y: y² - 8y = (y - 4)² - 16 Step 4: Substitute back: (x + 3)² - 9 + (y - 4)² - 16 = -9 Step 5: Simplify: (x + 3)² + (y - 4)² = 16 Step 6: Identify centre and radius: - Centre: (-3, 4) - Radius: √16 = 4
Example 2
Convert x² + y² - 4x + 2y - 4 = 0 to standard form
Solution:
Step 1: Group terms: (x² - 4x) + (y² + 2y) = 4 Step 2: Complete the square for x: x² - 4x = (x - 2)² - 4 Step 3: Complete the square for y: y² + 2y = (y + 1)² - 1 Step 4: Substitute back: (x - 2)² - 4 + (y + 1)² - 1 = 4 Step 5: Simplify: (x - 2)² + (y + 1)² = 9 Step 6: Identify centre and radius: - Centre: (2, -1) - Radius: √9 = 3
Geometric Applications
- Circles can be used in questions involving:
- Intersections with lines
- Tangents to the circle
- Geometry in coordinate systems
Example 1
Find the points of intersection between the circle (x - 2)² + (y - 1)² = 25 and the line y = 2x + 1
Solution:
Step 1: Substitute y = 2x + 1 into the circle equation: (x - 2)² + (2x + 1 - 1)² = 25 (x - 2)² + (2x)² = 25 Step 2: Expand and simplify: (x² - 4x + 4) + 4x² = 25 5x² - 4x + 4 = 25 5x² - 4x - 21 = 0 Step 3: Solve the quadratic equation: Using quadratic formula: x = (4 ± √(16 + 420)) / 10 x = (4 ± √436) / 10 x = (4 ± 2√109) / 10 x = (2 ± √109) / 5 Step 4: Find corresponding y values: y = 2x + 1 y = 2((2 ± √109) / 5) + 1 y = (4 ± 2√109) / 5 + 1 y = (9 ± 2√109) / 5 So the intersection points are: ((2 + √109)/5, (9 + 2√109)/5) and ((2 - √109)/5, (9 - 2√109)/5)
Example 2
Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4)
Solution:
Step 1: Verify that (3, 4) lies on the circle: 3² + 4² = 9 + 16 = 25 ✓ Step 2: Find the gradient of the radius from centre to point: Gradient of radius = (4 - 0) / (3 - 0) = 4/3 Step 3: The tangent is perpendicular to the radius: Gradient of tangent = -1 / (4/3) = -3/4 Step 4: Use point-gradient form: y - 4 = (-3/4)(x - 3) y - 4 = (-3/4)x + 9/4 y = (-3/4)x + 9/4 + 4 y = (-3/4)x + 25/4 So the equation of the tangent is y = (-3/4)x + 25/4
Advanced Circle Problems
More complex problems may involve multiple circles, circles and other curves, or optimization problems.
These often require combining circle geometry with other mathematical techniques.
Example 1
Two circles have equations (x - 2)² + (y - 3)² = 16 and (x + 1)² + (y - 1)² = 9. Find the distance between their centres and determine if they intersect.
Solution:
Step 1: Find the centres: Circle 1: centre (2, 3), radius 4 Circle 2: centre (-1, 1), radius 3 Step 2: Calculate distance between centres: d = √((2 - (-1))² + (3 - 1)²) d = √(3² + 2²) d = √(9 + 4) d = √13 ≈ 3.61 Step 3: Compare with radii: Sum of radii = 4 + 3 = 7 Difference of radii = 4 - 3 = 1 Since 1 < √13 < 7, the circles intersect at two points.
Example 2
Find the equation of the circle that passes through the points (1, 2), (3, 4), and (5, 2)
Solution:
Step 1: Use the general form (x - a)² + (y - b)² = r² Step 2: Substitute each point to get three equations: (1 - a)² + (2 - b)² = r² ... (1) (3 - a)² + (4 - b)² = r² ... (2) (5 - a)² + (2 - b)² = r² ... (3) Step 3: Expand and subtract equations to eliminate r²: From (1) and (2): (1 - a)² + (2 - b)² = (3 - a)² + (4 - b)² 1 - 2a + a² + 4 - 4b + b² = 9 - 6a + a² + 16 - 8b + b² -2a - 4b + 5 = -6a - 8b + 25 4a + 4b = 20 a + b = 5 ... (4) Step 4: From (1) and (3): (1 - a)² + (2 - b)² = (5 - a)² + (2 - b)² 1 - 2a + a² = 25 - 10a + a² -2a + 1 = -10a + 25 8a = 24 a = 3 Step 5: Substitute a = 3 into (4): 3 + b = 5, so b = 2 Step 6: Find r² using point (1, 2): r² = (1 - 3)² + (2 - 2)² = 4 + 0 = 4 So the equation is (x - 3)² + (y - 2)² = 4
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