Skip to main content
Chapter 21 of 31
Flashcards

Circles

Kerala Board · Class 12 · Mathematics

Flashcards for Circles — Kerala Board Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

44 questions20 flashcards5 concepts
20 Flashcards
Card 1Basic Definitions

What is the definition of a circle and what is its standard form equation?

Answer

A circle is the locus of a point which moves in a plane such that its distance from a fixed point remains constant. The fixed point is called the center and the constant distance is the radius. Stand

Card 2General Form

Convert the general form x² + y² + 2gx + 2fy + c = 0 to standard form and find center and radius.

Answer

General Form: x² + y² + 2gx + 2fy + c = 0 Step 1: Complete the square (x² + 2gx + g²) + (y² + 2fy + f²) = g² + f² - c Step 2: Standard form (x + g)² + (y + f)² = g² + f² - c Center: (-g, -f) Radius

Card 3Circle Equations

Find the equation of circle with center (2, -3) and radius 4.

Answer

Given: Center (h, k) = (2, -3), Radius r = 4 Step 1: Use standard form (x - h)² + (y - k)² = r² Step 2: Substitute values (x - 2)² + (y - (-3))² = 4² (x - 2)² + (y + 3)² = 16 Step 3: Expand to gene

Card 4Circle Equations

Find the center and radius of the circle x² + y² - 6x + 8y - 11 = 0

Answer

Given: x² + y² - 6x + 8y - 11 = 0 Step 1: Compare with x² + y² + 2gx + 2fy + c = 0 2g = -6 → g = -3 2f = 8 → f = 4 c = -11 Step 2: Find center and radius Center = (-g, -f) = (3, -4) Radius = √(g² +

Card 5Diameter Form

What is the equation of a circle when endpoints of a diameter are given as (x₁, y₁) and (x₂, y₂)?

Answer

Diameter Form of Circle: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0 Reason: If P(x, y) is any point on the circle, then angle APB = 90° (angle in semicircle) This means AP ⊥ BP, so slope of AP × slope o

Card 6Parametric Form

What are the parametric equations of a circle and when are they used?

Answer

Parametric Form of Circle: For circle with center (h, k) and radius r: x = h + r cos θ y = k + r sin θ where θ is the parameter (0 ≤ θ < 2π) For circle with center at origin: x = r cos θ, y = r sin θ

Card 7Circle Through Points

How do you find the equation of a circle passing through three non-collinear points?

Answer

Method: Use general form x² + y² + 2gx + 2fy + c = 0 Step 1: Substitute each point to get 3 equations Step 2: Solve for g, f, and c Example: Points (1, 1), (2, -1), (3, 2) Substituting: 1 + 1 + 2g

Card 8Point Position

How do you determine the position of a point P(x₁, y₁) with respect to circle S = x² + y² + 2gx + 2fy + c = 0?

Answer

Use S₁₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c Position of point P: • S₁₁ < 0: P is INSIDE the circle • S₁₁ = 0: P is ON the circle • S₁₁ > 0: P is OUTSIDE the circle Example: Point (2, 1) and circle x² + y²

+12 more flashcards available

Practice All

Get detailed flashcards for Circles

Super Tutor gives you interactive content for every chapter of Kerala Board Class 12 Mathematics — summaries, quizzes, flashcards, and more.

Try Super Tutor — It's Free

Frequently Asked Questions

What are the important topics in Circles for Kerala Board Class 12 Mathematics?

Circles covers several key topics that are frequently asked in Kerala Board Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.

Start by understanding all key concepts. Practise previous year questions from this chapter. Revise formulas and definitions regularly. Use flashcards for quick revision before the exam.

There are 20 flashcards for Circles covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.