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.

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

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.