Learn on PengienVision, Algebra 2Chapter 9: Conic Sections

Lesson 2: Circles

In this Grade 11 enVision Algebra 2 lesson, students learn to derive and apply the standard form of the equation of a circle, (x − h)² + (y − k)² = r², using the Pythagorean Theorem to connect geometric properties like center and radius to algebraic representations. Students practice writing and graphing circle equations, identifying domain and range, and solving real-world problems such as finding the center of a circular fence using the midpoint of a diameter.

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

Standard Form for a Circle

Property

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
For a circle centered at the origin (0,0)(0, 0), the equation simplifies to x2+y2=r2x^2 + y^2 = r^2.

Examples

  • A circle with center (4,−2)(4, -2) and radius 66 has the equation (x−4)2+(y−(−2))2=62(x - 4)^2 + (y - (-2))^2 = 6^2, which is (x−4)2+(y+2)2=36(x - 4)^2 + (y + 2)^2 = 36.
  • The equation (x+1)2+(y−5)2=81(x + 1)^2 + (y - 5)^2 = 81 describes a circle with center (−1,5)(-1, 5) and radius 81=9\sqrt{81} = 9.
  • The equation x2+y2=10x^2 + y^2 = 10 represents a circle centered at the origin (0,0)(0, 0) with a radius of 10\sqrt{10}.

Explanation

This equation is a direct application of the distance formula. It defines a circle as the set of all points (x,y)(x, y) that are exactly a distance rr from a center point (h,k)(h, k). This form makes the center and radius easy to identify.

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Chapter 9: Conic Sections

  1. Lesson 1

    Lesson 1: Parabolas

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  2. Lesson 2Current

    Lesson 2: Circles

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  3. Lesson 3

    Lesson 3: Ellipses

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  4. Lesson 4

    Lesson 4: Hyperbolas

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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

(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

Standard Form for a Circle

Property

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
For a circle centered at the origin (0,0)(0, 0), the equation simplifies to x2+y2=r2x^2 + y^2 = r^2.

Examples

  • A circle with center (4,−2)(4, -2) and radius 66 has the equation (x−4)2+(y−(−2))2=62(x - 4)^2 + (y - (-2))^2 = 6^2, which is (x−4)2+(y+2)2=36(x - 4)^2 + (y + 2)^2 = 36.
  • The equation (x+1)2+(y−5)2=81(x + 1)^2 + (y - 5)^2 = 81 describes a circle with center (−1,5)(-1, 5) and radius 81=9\sqrt{81} = 9.
  • The equation x2+y2=10x^2 + y^2 = 10 represents a circle centered at the origin (0,0)(0, 0) with a radius of 10\sqrt{10}.

Explanation

This equation is a direct application of the distance formula. It defines a circle as the set of all points (x,y)(x, y) that are exactly a distance rr from a center point (h,k)(h, k). This form makes the center and radius easy to identify.

Book overview

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

Continue this chapter

Chapter 9: Conic Sections

  1. Lesson 1

    Lesson 1: Parabolas

    LearnPractice
  2. Lesson 2Current

    Lesson 2: Circles

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  3. Lesson 3

    Lesson 3: Ellipses

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  4. Lesson 4

    Lesson 4: Hyperbolas

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