Geometry Notes G.12 Equations Of Circles Answer Key

Geometry Notes G.12: Equations of Circles – Answer Key

Introduction

Understanding the equations of circles is a cornerstone of Grade 12 geometry. Whether you are preparing for exams, solving real‑world problems, or building a foundation for advanced mathematics, mastering the standard and general forms of a circle’s equation will boost your confidence and performance. This guide provides clear notes, step‑by‑step procedures, and a comprehensive answer key to help you practice and verify your work Still holds up..

Standard Form of a Circle

The standard form directly reveals the circle’s center ((h, k)) and radius (r):

[ (x - h)^2 + (y - k)^2 = r^2 ]

• Center: ((h, k)) – the point about which the circle is symmetric.
• Radius: (r) – the distance from the center to any point on the circle.

Key point: If the equation is written exactly in this format, you can read off the center and radius instantly That's the part that actually makes a difference..

General Form of a Circle

The general form is a polynomial expansion:

[ x^2 + y^2 + Dx + Ey + F = 0 ]

To extract the center and radius, you must complete the square for both (x) and (y) terms.

Steps to convert general → standard:

1. Group (x)-terms and (y)-terms: ((x^2 + Dx) + (y^2 + Ey) = -F).
2. Complete the square for each group:
   • For (x): add (\left(\frac{D}{2}\right)^2).
   • For (y): add (\left(\frac{E}{2}\right)^2).
3. Balance the equation by adding the same numbers to the right‑hand side.
4. Factor the perfect squares to obtain ((x - h)^2 + (y - k)^2 = r^2).

Finding Center and Radius from Standard Form

• Center ((h, k)) is simply the numbers subtracted from (x) and (y).
• Radius (r) is the square root of the constant on the right side.

Example: ((x + 3)^2 + (y - 4)^2 = 25) → Center ((-3, 4)), Radius (\sqrt{25}=5) But it adds up..

Solving Circle Problems – Step‑by‑Step Process

1. Identify the given information (center, radius, points on the circle, or an equation).
2. Choose the appropriate form (standard or general) based on the problem.
3. Apply the relevant formula:
   • To find the equation from center and radius: plug into standard form.
   • To find the equation from three points: set up a system using the general form and solve for (D, E,) and (F).
4. Simplify and, if needed, convert between forms.
5. Check your answer by substituting known points or verifying the center‑radius relationship.

Answer Key – Sample Problems

Below are ten representative problems with detailed solutions. Use them to practice and self‑assess.

1. Problem: Write the equation of a circle with center ((2, -5)) and radius (4).
   Solution: Standard form → ((x - 2)^2 + (y + 5)^2 = 16) And that's really what it comes down to..

2. Problem: Convert (x^2 + y^2 - 6x + 8y - 11 = 0) to standard form.
   Solution:

   • Group: ((x^2 - 6x) + (y^2 + 8y) = 11)
   • Complete squares: ((x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16)
   • Factor: ((x - 3)^2 + (y + 4)^2 = 36)
   • Center ((3, -4)), Radius (6).

3. Problem: Find the center and radius of ((x + 1)^2 + (y - 7)^2 = 49).
   Solution: Center ((-1, 7)), Radius (\sqrt{49}=7).

4. Problem: Determine the equation of a circle passing through points ((1,2), (4,6), (5,3)).
   Solution:

   • Use general form: (x^2 + y^2 + Dx + Ey + F = 0).
   • Plug each point to create a system:
     [ \begin{cases} 1 + 4 + D(1) + E(2) + F = 0\ 16 + 36 + D(4) + E(6) + F = 0\ 25 + 9 + D(5) + E(3) + F = 0 \end{cases} ]
   • Solve (using elimination or matrix) → (D = -6, E = -2, F = 5).
   • Equation: (x^2 + y^2 - 6x - 2y + 5 = 0).

5. Problem: Write the general form of a circle with center ((-2, 3)) and radius (5).
   Solution: Start with standard ((x + 2)^2 + (y - 3)^2 = 25). Expand: (x^2 + 4x + 4 + y^2 - 6y + 9 = 25) → (x^2 + y^2 + 4x - 6y - 12 = 0) The details matter here..

6. Problem: Identify the type of conic represented by (x^2 + y^2 + 2x - 4y + 1 = 0).
   Solution: Complete squares → ((x + 1)^2 + (y - 2)^2 = 4). Since the right side is positive, it’s a circle.

7. Problem: Find the equation of a circle tangent to the line (y = -2) at the point ((3, -2)) and with center lying on the line (x = 5).
   Solution: Center must be ((5, k)). Tangency to (y = -2) means the radius is vertical, so (|k + 2| = \text{distance between center and point}). Since the point lies directly below the center, (k = -2). Thus center ((5, -2)) and radius (|5-3| = 2). Equation: ((x - 5)^2 + (y + 2)^2 = 4).

8. Problem: Convert ((