How to find Equation of a Circle

the equation of a circle provides an algebraic way to describe a circle and in this article, we teach you how to write two forms of the Equation of a Circle.

Rules for Equation of a Circle

• Equation of circles in standard form: $$(x- h)^2+( y-k)^2= r^2$$
• Equation of circles in general form: $$x^2+y^2+Ax+By+C=0$$

Examples

Equation of a Circle – Example 1:

Write the standard form equation of each circle.
$$x^2+ y^2-4x-6y+9=0$$

Solution:

The standard form of circle equation is: $$(x- h)^2+( y-k)^2= r^2$$
where the radius of the circle is $$r$$, and it’s centered at (h,k).
First, move the loose number to the right side: $$x^2+ y^2-4x-6y=-9$$
Group x-variables and y-variables together: $$(x^2-4x)+(y^2-6y)=-9$$
Convert $$x$$ to square form: $$(x^2-4x+4)+y^2-6y=-9+4→(x-2)^2+(y^2-6y)=-9+4$$
Convert $$y$$ to square form: $$(x-2)^2+(y^2-6y+9)=-9+4+9→(x-2)^2+(y-3)^2=4$$
Then:$$(x-2)^2+(y-3)^2=2^2$$

Equation of a Circle – Example 2:

Write the standard form equation of this circle.
$$x^2+ y^2+6x-10y-2=0$$

Solution:

$$(x- h)^2+( y-k)^2= r^2$$ is the circle equation with a radius r, centered at (h,k). To find this equation, first, move the loose number to the right side: $$x^2+ y^2+6x-10y=2$$
Group x-variables and y-variables together: $$(x^2+6x)+(y^2-10y)=2$$
Convert $$x$$ to square form:
$$(x^2+6x+9)+(y^2-10y)=2+9→(x+3)^2+(y^2-10y)=2+9→$$
Convert y to square form:
$$(x+3)^2+(y^2-10y+25)=2+9+25→(x+3)^2+(y-5)^2=36$$
Then: $$(x-(-3))^2+(y-5)^2=6^2$$

Equation of a Circle – Example 3:

Write the standard form equation of this circle.
$$x^2+ y^2+14x-12y+4=0$$

Solution:

$$(x- h)^2+( y-k)^2= r^2$$ is the circle equation with a radius r, centered at (h,k). To find this equation, first, move the loose number to the right side: $$x^2+ y^2+14x-12y+4=0$$
Group x-variables and y-variables together: $$(x^2+14x)+(y^2-12y)=-4$$
Convert $$x$$ to square form:
$$(x^2+14x+49)+(y^2-12y)=-4+49→(x+7)^2+(y^2-12y)=-4+49→$$
Convert y to square form:
$$(x+7)^2+(y^2-12y+36)=-4+49+36→(x+7)^2+(y-6)^2=81$$
Then: $$(x-(-7))^2+(y-6)^2=9^2$$

Equation of a Circle – Example 4:

Write the standard form equation of this circle.
$$x^2+ y^2+18x+2y-18=0$$

Solution:

$$(x- h)^2+( y-k)^2= r^2$$ is the circle equation with a radius r, centered at (h,k). To find this equation, first, move the loose number to the right side: $$x^2+ y^2+18x+2y=18$$
Group x-variables and y-variables together: $$(x^2+18x)+(y^2+2y)=18$$
Convert $$x$$ to square form:
$$(x^2+18x+81)+(y^2+2y)=18+81→(x+9)^2+(y^2+2y)=18+81→$$
Convert y to square form:
$$(x+9)^2+(y^2+2y+1)=18+81+1→(x+9)^2+(y+1)^2=100$$
Then: $$(x-(-9))^2+(y-(-1))^2=10^2$$

Exercises for Equation of a Circle

Write the standard form equation of each circle.

1. $$\color{blue}{x^2+y^2-2x+2y=7}$$
2. $$\color{blue}{x^2+y^2+12x+8y=-3}$$
3. $$\color{blue}{x^2+y^2+8x-2y=8}$$
4. $$\color{blue}{x^2+y^2-4x-8y-4=0}$$
5. Center: $$(2,-4)$$, Radius: 5
6. Center: $$(-1,-5)$$, Area: $$16π$$
1. $$\color{blue}{(x-1)^2+(y-(-1))^2=3^2}$$
2. $$\color{blue}{(x-(-6))^2+(y-(-4))^2=7^2}$$
3. $$\color{blue}{(x-(-4))^2+(y-1)^2=5^2}$$
4. $$\color{blue}{(x-2)^2+(y-4)^2=(2\sqrt{6})^2}$$
5. $$\color{blue}{(x-2)^2+(y-(-4))^2=5^2}$$
6. $$\color{blue}{(x-(-1))^2+(y-(-5))^2=4^2}$$

36% OFF

X

How Does It Work?

1. Find eBooks

Locate the eBook you wish to purchase by searching for the test or title.

3. Checkout

Complete the quick and easy checkout process.

Save up to 70% compared to print

Help save the environment