# How to Find Equation of a Circle? (+FREE Worksheet!)

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.

## Related Topics

- How to Find Arc Length and Sector Area
- How to Find the Area and Circumference of Circles
- How to Find the Center and the Radius of Circles

## Rules for Equation of a Circle

- Equation of circles in standard form: \((x- h)^2+( y-k)^2= r^2\)

Center: \(h,k\), Radius: \(r\) - Equation of circles in general form: \(x^2+y^2+Ax+By+C=0\)

**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.**

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

- \(\color{blue}{(x-1)^2+(y-(-1))^2=3^2}\)
- \(\color{blue}{(x-(-6))^2+(y-(-4))^2=7^2}\)
- \(\color{blue}{(x-(-4))^2+(y-1)^2=5^2}\)
- \(\color{blue}{(x-2)^2+(y-4)^2=(2\sqrt{6})^2}\)
- \(\color{blue}{(x-2)^2+(y-(-4))^2=5^2}\)
- \(\color{blue}{(x-(-1))^2+(y-(-5))^2=4^2}\)

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