Solved Exercises on Circles in Coordinate Plane in Algebra 2

Master circles in coordinate plane: equations, centers, radii, tangents, intersections, and problem-solving techniques through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Center-Radius Form
Exercise 1
Find the center and radius of the circle with equation (x - 2)² + (y + 3)² = 16, then graph the circle.
Definition:

Circle: The set of all points in a plane that are equidistant from a fixed point (the center)

Center-Radius Form: (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius

\((x - h)^2 + (y - k)^2 = r^2\)
Center-Radius Form of Circle
Solution Method:
  1. Compare the given equation to the standard form
  2. Identify h, k, and r from the equation
  3. Extract the center (h, k) and radius r
  4. Graph using center and radius
Given
(x-2)²+(y+3)²=16
Standard
(x-h)²+(y-k)²=r²
Result
Center: (2,-3), r=4
Step 1: Compare to standard form

(x - 2)² + (y + 3)² = 16

(x - h)² + (y - k)² = r²

Step 2: Identify parameters

h = 2 (from x - 2)

k = -3 (from y - (-3) = y + 3)

r² = 16, so r = 4

Step 3: State the answer

Center: (2, -3)

Radius: 4 units

Center: (2,-3), Radius: 4
Final answer:

Center: (2, -3), Radius: 4

Applied rules:

• Pattern matching: Compare coefficients to identify parameters

• Sign handling: Be careful with signs in the standard form

• Radical simplification: Take positive square root for radius

Tip 1: The center coordinates are the opposite of the numbers in parentheses.
Tip 2: Radius is always positive, so take the positive square root.
2 General Form Conversion
Exercise 2
Convert the equation x² + y² - 6x + 8y + 9 = 0 to center-radius form and identify the circle's center and radius.
Definition:

General Form: x² + y² + Dx + Ey + F = 0

Completing the Square: A technique to convert general form to center-radius form

\(x^2 + y^2 + Dx + Ey + F = 0 \rightarrow (x-h)^2 + (y-k)^2 = r^2\)
Conversion Process
Original
x²+y²-6x+8y+9=0
Complete squares
(x-3)²+(y+4)²=16
Result
Center: (3,-4), r=4
Step 1: Group x and y terms

x² - 6x + y² + 8y + 9 = 0

(x² - 6x) + (y² + 8y) = -9

Step 2: Complete the square for x terms

Take half of coefficient of x: -6/2 = -3

Square it: (-3)² = 9

Add to both sides: (x² - 6x + 9)

Step 3: Complete the square for y terms

Take half of coefficient of y: 8/2 = 4

Square it: 4² = 16

Add to both sides: (y² + 8y + 16)

Step 4: Factor and simplify

(x² - 6x + 9) + (y² + 8y + 16) = -9 + 9 + 16

(x - 3)² + (y + 4)² = 16

Step 5: Identify center and radius

Center: (3, -4)

Radius: √16 = 4

Center: (3,-4), Radius: 4
Final answer:

Center-radius form: (x - 3)² + (y + 4)² = 16, Center: (3, -4), Radius: 4

Applied rules:

• Completing the square: Add (coefficient/2)² to both sides

• Perfect square trinomial: x² + bx + (b/2)² = (x + b/2)²

• Standard form recognition: Match to center-radius form

Tip 1: Always add the same amount to both sides when completing the square.
Tip 2: Work with x and y terms separately when completing the square.
3 Point on Circle
Exercise 3
Determine if the point (5, -1) lies inside, outside, or on the circle with equation (x - 2)² + (y + 3)² = 16.
Definition:

Point Position: Compare distance from point to center with radius

Distance Formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]

\(\text{If } (x-a)^2 + (y-b)^2 \begin{cases} < r^2 & \text{point inside} \\ = r^2 & \text{point on circle} \\ > r^2 & \text{point outside} \end{cases}\)
Point Position Relative to Circle
Given
Point: (5,-1)
Circle
(x-2)²+(y+3)²=16
Result
Point inside circle
Step 1: Identify circle parameters

Center: (2, -3)

Radius: r = √16 = 4

Step 2: Substitute point into circle equation

(5 - 2)² + (-1 + 3)² = ?

(3)² + (2)² = 9 + 4 = 13

Step 3: Compare with radius squared

Since 13 < 16, the point lies inside the circle

Step 4: Verify with distance formula

Distance from (5, -1) to (2, -3):

d = √[(5-2)² + (-1-(-3))²] = √[9 + 4] = √13 ≈ 3.6

Since 3.6 < 4, point is inside the circle

Point (5,-1) lies inside the circle
Final answer:

The point (5, -1) lies inside the circle.

Applied rules:

• Substitution test: Plug point into circle equation

• Comparison: Compare result with radius squared

• Distance verification: Use distance formula to confirm

Tip 1: Substitution is often faster than calculating distance.
Tip 2: Use distance formula when you need exact distance to center.
Key Formulas and Properties
\((x - h)^2 + (y - k)^2 = r^2\)
Center-Radius Form
\(x^2 + y^2 + Dx + Ey + F = 0\)
General Form
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Distance Formula
Position Test
Point vs Circle
(x-a)²+(y-b)² vs r²
Tangent Line
Perpendicular to radius
m_tangent × m_radius = -1
Intersection
Circle-Line System
Solve simultaneous equations
Key Definitions:

Circle: Set of all points equidistant from a fixed point (center)

Center: Fixed point from which all points on circle are equidistant

Radius: Distance from center to any point on circle

Chord: Line segment connecting two points on circle

Diameter: Chord passing through center

Problem-Solving Strategy:
  1. Identify form: Recognize center-radius or general form
  2. Convert if needed: Complete the square to find center and radius
  3. Extract parameters: Identify center (h, k) and radius r
  4. Apply relevant formulas: Use distance formula, position tests, etc.
  5. Verify results: Check that answers make sense geometrically
Common Errors: Sign errors in center coordinates, incorrect radius calculation, forgetting to complete the square properly.
Exam Tips: Memorize both forms, practice completing the square, understand parameter relationships.
Solution: Exercises 4 to 5
4 Tangent Line
Exercise 4
Find the equation of the tangent line to the circle (x - 1)² + (y - 2)² = 9 at the point (4, 2).
Definition:

Tangent Line: A line that touches the circle at exactly one point

Perpendicular Property: Tangent line is perpendicular to the radius at the point of tangency

Given
Circle: (x-1)²+(y-2)²=9
Point
(4,2)
Result
Tangent: x=4
Step 1: Verify point is on circle

(4 - 1)² + (2 - 2)² = 3² + 0² = 9 ✓

Step 2: Find center and radius

Center: (1, 2)

Radius: √9 = 3

Step 3: Find slope of radius

From center (1, 2) to point (4, 2):

m_radius = (2 - 2)/(4 - 1) = 0/3 = 0

Step 4: Find slope of tangent

Since radius has slope 0 (horizontal), tangent is vertical

For perpendicular lines: m_tangent × m_radius = -1

Since m_radius = 0, the tangent line is vertical

Step 5: Write equation of tangent

Vertical line through (4, 2): x = 4

Equation of tangent: x = 4
Final answer:

The equation of the tangent line is x = 4.

Applied rules:

• Tangent-perpendicularity: Tangent line ⊥ radius at point of tangency

• Slope relationship: m₁ × m₂ = -1 for perpendicular lines

• Special case: Horizontal radius gives vertical tangent

Tip 1: Always verify that the point lies on the circle first.
Tip 2: Remember that vertical lines have undefined slope and equation x = constant.
5 Circle Intersection
Exercise 5
Find the intersection points of the circles (x - 1)² + (y - 1)² = 4 and (x - 4)² + (y - 1)² = 1.
Definition:

Circle Intersection: Points where two circles meet simultaneously

Simultaneous Equations: Solve system of two circle equations

Given
C₁: (x-1)²+(y-1)²=4
C₂: (x-4)²+(y-1)²=1
Method
Subtraction eliminates y²
Result
Intersection: (3,1±√3)
Step 1: Write both equations

Circle 1: (x - 1)² + (y - 1)² = 4

Circle 2: (x - 4)² + (y - 1)² = 1

Step 2: Subtract equations to eliminate y²

[(x - 1)² + (y - 1)²] - [(x - 4)² + (y - 1)²] = 4 - 1

(x - 1)² - (x - 4)² = 3

Step 3: Expand and simplify

(x² - 2x + 1) - (x² - 8x + 16) = 3

x² - 2x + 1 - x² + 8x - 16 = 3

6x - 15 = 3

6x = 18

x = 3

Step 4: Substitute back to find y

Substitute x = 3 into Circle 1:

(3 - 1)² + (y - 1)² = 4

4 + (y - 1)² = 4

(y - 1)² = 0

y - 1 = 0

y = 1

Step 5: Verify with Circle 2

(3 - 4)² + (1 - 1)² = 1 + 0 = 1 ✓

Intersection point: (3, 1)
Final answer:

The circles intersect at the point (3, 1).

Applied rules:

• System solving: Use elimination method to solve simultaneous equations

• Algebraic manipulation: Expand and simplify expressions

• Verification: Always check solutions in both original equations

Tip 1: Subtracting equations often eliminates quadratic terms.
Tip 2: Always verify intersection points in both original equations.
Comprehensive Guide: Circles in Coordinate Plane
\((x - h)^2 + (y - k)^2 = r^2\)
Center-Radius Form
\(x^2 + y^2 + Dx + Ey + F = 0\)
General Form
Key definitions:

Circle: The set of all points in a plane that are equidistant from a fixed point called the center

Center: The fixed point from which all points on the circle are equidistant

Radius: The distance from the center to any point on the circle

Chord: A line segment connecting two points on the circle

Diameter: A chord that passes through the center of the circle

Complete methodology:
  1. Identify the form: Recognize center-radius or general form of the equation
  2. Convert if needed: Complete the square to transform general form to center-radius form
  3. Extract parameters: Identify center (h, k) and radius r from center-radius form
  4. Apply formulas: Use distance formula, point-position tests, or other relevant formulas
  5. Verify results: Check that answers satisfy the original equation
Tip 1: The center coordinates are the opposite of the numbers in parentheses in center-radius form.
Tip 2: Radius is always positive, so take the positive square root of r².
Tip 3: For point position tests, compare (x-h)² + (y-k)² with r².
Tip 4: The tangent line at any point on a circle is perpendicular to the radius at that point.
Common errors: Sign errors in center coordinates, incorrect radius calculation, forgetting to complete the square properly, calculation errors in distance formula.
Exam preparation: Memorize both forms, practice completing the square, work on tangent line problems, study circle-circle intersections.
Essential properties to know:

• Center-radius form: (x - h)² + (y - k)² = r²

• General form: x² + y² + Dx + Ey + F = 0

• Conversion: Complete the square for x and y terms separately

• Point position: (x - h)² + (y - k)² <, =, or > r² determines inside/on/outside

Visual Understanding: Circle Properties
Exercise 6: Circle Comparison
Compare different circles:
• (x - 0)² + (y - 0)² = 1 (unit circle centered at origin)
• (x - 2)² + (y - 3)² = 4 (radius 2, center at (2,3))
• (x + 1)² + (y - 2)² = 9 (radius 3, center at (-1,2))
• (x - 1)² + (y + 1)² = 16 (radius 4, center at (1,-1))

Analysis: The visualization shows how different centers and radii affect circle positions and sizes.

  • Center coordinates determine position in the plane
  • Radius determines the size of the circle
  • Larger radii create bigger circles
  • Changing center moves the entire circle

Questions & Answers

Question: How do I convert from general form to center-radius form? The completing the square part confuses me.

Answer: Here's the systematic approach for completing the square:

  1. Group terms: Put x terms together and y terms together
  2. Move constant: Move the constant term to the other side of the equation
  3. Complete for x: Take half of the x-coefficient, square it, add to both sides
  4. Complete for y: Take half of the y-coefficient, square it, add to both sides
  5. Factor: Write as perfect squares

For example, with x² + y² - 6x + 8y + 9 = 0:
1. Group: (x² - 6x) + (y² + 8y) = -9
2. Complete x: (-6/2)² = 9, so (x² - 6x + 9)
3. Complete y: (8/2)² = 16, so (y² + 8y + 16)
4. Add to both sides: (x² - 6x + 9) + (y² + 8y + 16) = -9 + 9 + 16
5. Factor: (x - 3)² + (y + 4)² = 16

Always remember to add the same amount to both sides!

Question: How do I find the equation of a tangent line to a circle at a given point?

Answer: The key is the perpendicular property of tangent lines:

  1. Find the center: From the circle equation
  2. Calculate radius slope: Slope from center to the point of tangency
  3. Find tangent slope: Use m_tangent = -1/m_radius
  4. Write equation: Use point-slope form with the given point and tangent slope

For example, for circle (x - 2)² + (y - 3)² = 9 at point (5, 3):
1. Center: (2, 3)
2. Radius slope: (3-3)/(5-2) = 0/3 = 0 (horizontal line)
3. Tangent slope: undefined (perpendicular to horizontal = vertical)
4. Equation: x = 5 (vertical line through (5, 3))

Special case: If radius slope is 0, tangent is vertical; if radius slope is undefined, tangent is horizontal.

Question: What are some real-world applications of circles in coordinate geometry? Why is it important to learn this?

Answer: Circles in coordinate geometry have numerous practical applications:

  • Engineering: Wheel design, gear systems, circular motion mechanics
  • Physics: Orbital mechanics, pendulum motion, wave propagation
  • Computer Graphics: Rendering circular objects, collision detection
  • Architecture: Dome construction, round building design
  • Navigation: GPS positioning, radar systems

Learning circles in coordinate geometry develops understanding of curved geometric shapes and their properties. This knowledge is fundamental to trigonometry, calculus, and physics. The mathematical techniques used (distance formula, completing the square) are essential for higher mathematics.

Understanding circles helps in modeling real-world phenomena and solving practical problems in science and engineering.

Question: I sometimes mix up the center coordinates when reading the equation. Any tips to avoid this mistake?

Answer: Here are strategies to remember the center coordinates:

  1. Sign flip rule: The center coordinates are the OPPOSITE of the numbers in parentheses
  2. Pattern recognition: (x - h)² + (y - k)² = r² means center is (h, k)
  3. Verification: Substitute center coordinates into the equation to check
  4. Memory aid: "What makes the expression zero?" - if (x - 3)², then x = 3 makes it zero

For example, in (x - 5)² + (y + 2)² = 9:
• From (x - 5)²: h = 5 (opposite of -5)
• From (y + 2)²: k = -2 (opposite of +2)
• Center: (5, -2)

Always verify by checking that (h, k) satisfies the equation: (5-5)² + (-2+2)² = 0 + 0 = 0 ≠ r², but this point is the center where the expression is minimized.

Question: How do I determine the number of intersection points between two circles?

Answer: The number of intersection points depends on the distance between centers and the radii:

  • 0 intersections: Distance between centers > r₁ + r₂ (circles are too far apart)
  • 0 intersections: Distance between centers < |r₁ - r₂| (one circle is inside the other)
  • 1 intersection: Distance between centers = r₁ + r₂ (externally tangent)
  • 1 intersection: Distance between centers = |r₁ - r₂| (internally tangent)
  • 2 intersections: |r₁ - r₂| < Distance between centers < r₁ + r₂ (circles intersect)

To find the actual intersection points, solve the system of equations by subtracting one equation from the other to eliminate the quadratic terms, then substitute back.

The algebraic solution will match the geometric possibilities.