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

(x−h)2+(y−k)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 (x−3)2+(y−(−1))2=52(x - 3)^2 + (y - (-1))^2 = 5^2, which simplifies to (x−3)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

Lesson overview

Expand to review the lesson summary and core properties.

Expand

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

(x−h)2+(y−k)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 (x−3)2+(y−(−1))2=52(x - 3)^2 + (y - (-1))^2 = 5^2, which simplifies to (x−3)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