Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 14: Graphing Quadratics

Lesson 2: Circles

In this lesson from AoPS Introduction to Algebra, Grade 4 students learn how to derive and apply the standard form equation of a circle, (x − h)² + (y − k)² = r², where (h, k) is the center and r is the radius. Students use the distance formula to understand why a circle is defined as all points equidistant from a center, and practice identifying centers and radii from equations, as well as converting non-standard equations into standard form by completing the square and dividing to normalize coefficients.

Section 1

Circle

Property

A circle is the set of all points in a plane that lie at a given distance, called the radius, from a fixed point called the center. The equation for a circle of radius rr centered at the point (h,k)(h, k) is

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

Examples

  • The equation for a circle with its center at (3,1)(3, -1) and a radius of 55 is (x3)2+(y(1))2=52(x - 3)^2 + (y - (-1))^2 = 5^2, which simplifies to (x3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25.

Section 2

Completing the Square for Circle Equations

Property

To convert a circle equation from general form to standard form, we complete the square for both xx and yy terms. For an expression x2+bxx^2 + bx, we add and subtract the constant (b2)2(\frac{b}{2})^2 to create a perfect square.

  1. Group the xx terms and yy terms separately
  2. For x2+bxx^2 + bx, add and subtract (b2)2(\frac{b}{2})^2
  3. For y2+dyy^2 + dy, add and subtract (d2)2(\frac{d}{2})^2
  4. Factor each group: x2+bx+(b2)2=(x+b2)2x^2 + bx + (\frac{b}{2})^2 = (x + \frac{b}{2})^2

Section 3

Finding Circle Equations from Three Points

Property

To find a circle equation through three non-collinear points, substitute each point into the general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 to create a system of three linear equations in variables DD, EE, and FF.

Examples

Book overview

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Chapter 14: Graphing Quadratics

  1. Lesson 1

    Lesson 1: Parabolas

  2. Lesson 2Current

    Lesson 2: Circles

Lesson overview

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Section 1

Circle

Property

A circle is the set of all points in a plane that lie at a given distance, called the radius, from a fixed point called the center. The equation for a circle of radius rr centered at the point (h,k)(h, k) is

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

Examples

  • The equation for a circle with its center at (3,1)(3, -1) and a radius of 55 is (x3)2+(y(1))2=52(x - 3)^2 + (y - (-1))^2 = 5^2, which simplifies to (x3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25.

Section 2

Completing the Square for Circle Equations

Property

To convert a circle equation from general form to standard form, we complete the square for both xx and yy terms. For an expression x2+bxx^2 + bx, we add and subtract the constant (b2)2(\frac{b}{2})^2 to create a perfect square.

  1. Group the xx terms and yy terms separately
  2. For x2+bxx^2 + bx, add and subtract (b2)2(\frac{b}{2})^2
  3. For y2+dyy^2 + dy, add and subtract (d2)2(\frac{d}{2})^2
  4. Factor each group: x2+bx+(b2)2=(x+b2)2x^2 + bx + (\frac{b}{2})^2 = (x + \frac{b}{2})^2

Section 3

Finding Circle Equations from Three Points

Property

To find a circle equation through three non-collinear points, substitute each point into the general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 to create a system of three linear equations in variables DD, EE, and FF.

Examples

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 14: Graphing Quadratics

  1. Lesson 1

    Lesson 1: Parabolas

  2. Lesson 2Current

    Lesson 2: Circles