Circle

Circle#

Distance Formula Revision#

Theory#

The equation of a circle relies on the distance formula:

  • Distance from point \((x, y)\) to center \((h, k)\) is: \(d = \sqrt{(x - h)^2 + (y - k)^2}\)

  • A circle is the set of all points at a fixed distance (radius) from a center

  • This fundamental concept leads to the circle equation

Application#

All points at distance 5 from center \((2, 3)\) satisfy: \(\sqrt{(x - 2)^2 + (y - 3)^2} = 5\)

Equation of a Circle#

Theory#

A circle can be represented in different forms:

Standard Form#

A circle with center \((h, k)\) and radius \(r\) has equation: $\((x - h)^2 + (y - k)^2 = r^2\)$

General Form#

Expanding the standard form gives: $\(x^2 + y^2 + 2gx + 2fy + c = 0\)$

where:

  • Center: \((-g, -f)\)

  • Radius: \(r = \sqrt{g^2 + f^2 - c}\) (exists only if \(g^2 + f^2 - c > 0\))

Special Cases#

  1. Circle centered at origin: \(x^2 + y^2 = r^2\)

  2. Unit circle: \(x^2 + y^2 = 1\) (center at origin, radius 1)

  3. Point circle: When \(r = 0\) (degenerate case)

Interactive Visualization: Circle Equation Explorer#

Application#

Examples#

Example 1: Finding Center and Radius Find the center and radius of the circle \(x^2 + y^2 - 6x + 8y - 11 = 0\).

Solution: Rearrange and complete the square: \((x^2 - 6x) + (y^2 + 8y) = 11\) \((x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16\) \((x - 3)^2 + (y + 4)^2 = 36\)

Therefore:

  • Center: \((3, -4)\)

  • Radius: \(r = \sqrt{36} = 6\)

Example 2: Writing Circle Equation Write the equation of a circle with center \((2, -3)\) passing through point \((5, 1)\).

Solution: First, find the radius using the distance formula: \(r = \sqrt{(5 - 2)^2 + (1 - (-3))^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

The equation is: \((x - 2)^2 + (y + 3)^2 = 25\)

Expanding to general form: \(x^2 + y^2 - 4x + 6y - 12 = 0\)

Example 3: Circle Through Three Points Find the equation of the circle passing through \((1, 1)\), \((2, 4)\), and \((5, 3)\).

Solution: Let the general equation be \(x^2 + y^2 + 2gx + 2fy + c = 0\).

Substituting the three points:

  • \((1, 1)\): \(1 + 1 + 2g + 2f + c = 0\) → \(2g + 2f + c = -2\)

  • \((2, 4)\): \(4 + 16 + 4g + 8f + c = 0\) → \(4g + 8f + c = -20\)

  • \((5, 3)\): \(25 + 9 + 10g + 6f + c = 0\) → \(10g + 6f + c = -34\)

Solving this system: \(g = -3\), \(f = -2\), \(c = 12\)

Therefore: \(x^2 + y^2 - 6x - 4y + 12 = 0\)

Multiple Choice Questions#

Sector Specific Questions: Circle Applications#

Key Takeaways#

Important

Circle Equations - Essential Concepts

  1. Standard Form: \((x - h)^2 + (y - k)^2 = r^2\)

    • Center: \((h, k)\)

    • Radius: \(r\)

  2. General Form: \(x^2 + y^2 + 2gx + 2fy + c = 0\)

    • Center: \((-g, -f)\)

    • Radius: \(r = \sqrt{g^2 + f^2 - c}\) (if \(g^2 + f^2 - c > 0\))

  3. Key Techniques:

    • Complete the square to convert between forms

    • Use distance formula to find radius

    • Three non-collinear points determine a unique circle

  4. Applications: Radar coverage, particle physics, delivery zones, lighting design

  5. Remember: A circle exists only when \(g^2 + f^2 - c > 0\) in general form

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