In this mini-lesson, we will learn about the equation of a circle by understanding the different ways of representing a circle and the ways to represent a circle on a graph. We will also discover interesting facts around the equation of a circle.

Joy is playing with a rope in a playground. He ties the rope to a point on the ground. He ties a stick to the other free end of the rope.

He stretches the rope as much as he can and starts moving with the rope and stick, such that the stick marks the path. He realizes after some time that he has reached the same point from where he had started. He also realizes that the path traced by the stick, as he is moving at a fixed distance from a point, is nothing but a circle.

Have you ever come across the equation of a circle when the coordinates of the center and the radius are given?

Play with the simulation below and explore the different equations of a circle.

**Lesson Plan**

**What Is a Circle?**

A **circle** is a figure obtained by joining all the points in a plane such that every point is at an equal distance from the point at the center.

**Radius**

Radius of a circle is the distance between the center and any point on the boundary of the circle.

**Circle on a Graph**

In the figure above, we can see a circle on a graph.

The center of the circle is point \((x_1, y_1)\), and the points \((x_2, y_2)\) and \((x_3, y_3)\) are the two points on the circumference of the circle.

Radius \((r)\) of the circle is the distance between the center and any one of the two points.

\( r = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\) or \( r = \sqrt{(y_3 - y_1)^2 + (x_3 - x_1)^2}\)

**What Is the Standard Form of Equation of a Circle?**

\((x_1, y_1)\) is the center of the circle with radius \(r\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - x_1)^2 + (y - y_1)^2} = r\)

Squaring both sides, we get:

\( (x - x_1)^2 + (y - y_1)^2 = r^2\)

This is the standard form of the equation of a circle.

Standard form \( (x - x_1)^2 + (y - y_1)^2 = r^2\) |

**Example**

Write the equation of a circle with center \((-1, 2)\) and radius equal to 3 units in the standard form.

**Solution**

Equation for a circle in standard form is given as:

\( (x - x_1)^2 + (y - y_1)^2 = r^2\)

where \((x_1, y_1) = (-1, 2)\) is the center of the circle and \(r = 3\) is the radius of the circle.

Substituting the values in the equation

\( \begin{align}

(x - (-1))^2 + (y - 2)^2 &= 3^2\\[0.2cm]

(x + 1)^2 + (y - 2)^2 &= 9\\[0.2cm]

\end{align}\)

\( (x + 1)^2 + (y - 2)^2 = 9\) |

**Center at Origin**

In the above figure, \((0, 0)\) is the center of the circle with radius \(r\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - 0)^2 + (y - 0)^2} = r\)

Squaring both sides, we get:

\( (x - 0)^2 + (y - 0)^2 = r^2\)

\( x^2 + y^2 = r^2\)

\( x^2 + y^2 = r^2\) |

**Center on X-axis**

\((a, 0)\) is the center of the circle with radius \(r\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - a)^2 + (y - 0)^2} = r\)

Squaring both sides, we get:

\( (x - a)^2 + (y - 0)^2 = r^2\)

\( (x - a)^2 + (y)^2 = r^2\)

\( (x - a)^2 + y^2 = r^2\) |

**Center on Y-axis**

\((0, b)\) is the center of the circle with radius \(r\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - 0)^2 + (y - b)^2} = r\)

Squaring both sides, we get:

\( (x - 0)^2 + (y - b)^2 = r^2\)

\( (x)^2 + (y - b)^2 = r^2\)

\( x^2 + (y - b)^2 = r^2\) |

**Circle Touching X-axis**

\((a, r)\) is the center of the circle with radius \(r\).

If a circle touches the X-axis, then the y-coordinate of the center of the circle is equal to the radius \(r\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - a)^2 + (y - r)^2} = r\)

Squaring both sides, we get:

\( (x - a)^2 + (y - r)^2 = r^2\)

\( (x - a)^2 + (y - r)^2 = r^2\) |

**Circle Touching Y-axis**

\((r, b)\) is the center of the circle with radius \(r\).

If a circle touches the Y-axis, then the x-coordinate of the center of the circle is equal to the radius \(r\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - r)^2 + (y - b)^2} = r\)

Squaring both sides

\( (x - r)^2 + (y - b)^2 = r^2\)

\( (x - r)^2 + (y - b)^2 = r^2\) |

**Circle Touching Both X-axis and Y-axis**

\((r, r)\) is the center of the circle with radius \(r\).

If a circle touches both X-axis and Y-axis, then the coordinates of the center of the circle is equal to the radius \((r, r)\).

\((x, y)\) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle.

Let's apply the distance formula between these points.

\( \sqrt{(x - r)^2 + (y - r)^2} = r\)

Squaring both sides

\( (x - r)^2 + (y - r)^2 = r^2\)

\( (x - r)^2 + (y - r)^2 = r^2\) |

**If a circle touches both the axes, then consider the center of the circle to be \((r, r)\), where \(r\) is the radius of the circle. Here, \((r, r)\) can be positive as well as negative.**

For example, the radius of the circle is 3 and it is touching both the axes, then the coordinates of the center can be \((3, 3)\), \((3, -3)\), \((-3, 3)\), or \((-3, -3)\).**A tangent touches a circle at one point. Hence, the point of contact of the tangent satisfies the equation of the circle as well as the equation of the tangent.**

**Conversion From Standard Form to the General Form**

Standard form of the equation of a circle is:

\( (x - x_1)^2 + (y - y_1)^2 = r^2\)

Let's expand the above equation using the identities.

\( x^2 +{x_1}^2 -2xx_1 + y^2 +{y_1}^2 -2yy_1 = r^2\)

\( x^2 + y^2 - 2xx_1 - 2yy_1 + {x_1}^2 + {y_1}^2 = r^2\)

\( x^2 + y^2 - 2xx_1 - 2yy_1 + {x_1}^2 + {y_1}^2 -r^2 = 0\)

Replace \(-2x_1\) with \(A\), \(-2y_1\) with \(B\), \( {x_1}^2 + {y_1}^2 -r^2\) with \(C\)

\( x^2 + y^2 + Ax + By + C = 0\)

General form of equation of circle is:

General form \( x^2 + y^2 + Ax + By + C = 0\) |

Where A, B, C are constants

- The general form of equation of circle always has \(x^2 + y^2\) in the beginning.
- If any equation is of the form \(x^2 +y^2 +axy + C = 0\), then it is not the equation of the circle. There is no \(xy\) term in the equation of a circle.
- If a circle crosses both the axes, then there are four points of intersection of the circle and the axes.
- If a circle touches both the axes, then there are only two points of contact.
- Radius is the distance from the center to any point on the boundary of the circle. Hence, the value of the radius of circle is always positive.

Now that you have understood about the equation of a circle, you can explore this simulation to see some more equations of circle by putting the value of the coordinates of the center and the radius of the circle.

**Solved Examples**

Example 1 |

Gerry's friend gave him an equation of a circle \( x^2 + y^2 +6x + 8y + 9 = 0\) and asked him to find the coordinates of the center and the radius of the circle. Help him solve the problem.

**Solution**

Given the equation of the circle \( x^2 + y^2 +6x + 8y + 9 = 0\)

The general form of the equation of the circle with center \((x_1, y_1)\) and radius \(r\) is \( x^2 + y^2 + Ax + By + C = 0\)

where \(A = -2x_1\)

\(B = -2y_1\)

\(C = {x_1}^2 + {y_1}^2 -r^2\)

From the equation of the circle \( x^2 + y^2 +6x + 8y + 9 = 0\)

\[\begin{align}

A &= 6 \\[0.2cm]

-2x_1 &= 6 \\[0.2cm]

x_1 &= -3 \\[0.2cm]

B &= 8 \\[0.2cm]

-2y_1 &= 8 \\[0.2cm]

y_1 &= -4 \\[0.2cm]

C &= 9 \\[0.2cm]

{x_1}^2 + {y_1}^2 -r^2 &= 9 \\[0.2cm]

{-3}^2 + {-4}^2 -r^2 &= 9 \\[0.2cm]

9 + 16 -r^2 &= 9 \\[0.2cm]

r^2 &= 16 \\[0.2cm]

r &= 4 \\[0.2cm]

\end{align}\]

\(\therefore\) Center of the circle is (-3, -4) and radius is 4 units |

Example 2 |

Find the point of intersection of the circle \(x^2 + y^2 -7x -8y +12 = 0\) and the axes.

**Solution**

If the circle crosses the X-axis, the value of the y-coordinate is zero

Substitute y = 0 in the equation of circle

\[\begin{align}

x^2 + y^2 -7x -8y +12 &= 0 \\[0.2cm]

x^2 + 0^2 -7x -8 \times 0 +12 &= 0 \\[0.2cm]

x^2 -7x +12 &= 0 \\[0.2cm]

(x -3)(x-4) &= 0 \\[0.2cm]

x &= 3 \\[0.2cm]

x &= 4 \\[0.2cm]

\end{align}\]

Hence, the circle crosses the x-axis at \((3, 0), (4, 0)\).

If the circle crosses the Y-axis, the value of the x-coordinate is zero.

Substitute x = 0 in the equation of circle

\[\begin{align}

x^2 + y^2 -7x -8y +12 &= 0 \\[0.2cm]

0^2 + y^2 -7 \times 0 - 8y +12 &= 0 \\[0.2cm]

y^2 - 8y +12 &= 0 \\[0.2cm]

(y - 6)(y - 2) &= 0 \\[0.2cm]

y &= 6 \\[0.2cm]

y &= 2 \\[0.2cm]

\end{align}\]

Hence, the circle crosses the y-axis at \((0, 2), (0, 6)\).

\(\therefore\) The circle crosses the axes at \((3, 0), (4, 0), (0, 2), (0, 6)\) |

Example 3 |

The line \(y + x = 1\) and the circle \({(x - 1)}^2 + {(y - 1)}^2 = 1\) intersect at two points. Find the coordinates of these points.

**Solution**

To find the coordinates of the points of intersection of the line and the circle, let's solve these two equations simultaneously.

Equation of line: \(y + x = 1\)

Equation of the circle: \({(x - 1)}^2 + {(y - 1)}^2 = 1\)

Put \(y = 1 -x\) in the equation of circle

\[\begin{align}

{(x - 1)}^2 + {(1 - x - 1)}^2 &= 1 \\[0.2cm]

{(x - 1)}^2 + {(- x)}^2 &= 1 \\[0.2cm]

{(x - 1)}^2 + x^2 &= 1 \\[0.2cm]

{(x - 1)}^2 &= 1 - x^2 \\[0.2cm]

{(x - 1)}^2 &= (1 - x)(1 + x) \\[0.2cm]

{(x - 1)}^2 - (1 - x)(1 + x) &= 0\\[0.2cm]

{(x - 1)}^2 + (x - 1)(1 + x) &= 0\\[0.2cm]

(x - 1)(x - 1 + 1 + x) &= 0\\[0.2cm]

(x - 1)(2x) &= 0\\[0.2cm]

x - 1 &= 0\\[0.2cm]

2x &= 0\\[0.2cm]

\end{align}\]

\[x = 1 \text{ and } x = 0\]

Substituting these value of x in the equation of line

\[y + 1 = 1\]

\[y = 0\]

\[y + 0 = 1\]

\[y = 1\]

Hence, \((1, 0), (0, 1)\) are the two points of intersection of the line and the circle.

\(\therefore\) (1, 0), (0, 1) |

**Interactive Questions**

**Here are a few activities for you to practice. **

**Select/Type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

This mini-lesson helped you understand the fascinating concept of the equation of a circle. The math journey around equation of circle starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that is not only relatable but also easy to grasp and will stay with them forever. Here lies the magic with Cuemath.

**About Cuemath**

At Cuemath, our team of math experts are dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it simulations, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in.

**Frequently Asked Questions (FAQs)**

## 1. What is the general equation of a circle?

The general equation of the circle is

\((x-h)^{2} +(y-k)^{2} = r^2\)

Where \((h,k)\) are the coordinates of the center and \(r\) is the length of the radius of the circle.

## 2. How do you write the standard form of a circle?

The standard form of equation of circle is

\(x^2 + y^2 +2gx + 2hy + c = 0\)

Center of the circle is \((-g, -h)\) and radius of the circle is \(r = g^2 + h^2 - c\)