Table of Contents

# How to find the Equation of a circle

Here we will learn about all the rules to find the equation of a circle with examples. But at first, we just remind ourselves what a circle is?

The locus of a moving point which is always at the same distance from a fixed point is called a circle. The fixed point is known as the center of the circle and the distance from that fixed point and the moving point is called radius.

## Important rules in equation of circle

1. The equation of a circle whose center is at the origin i.e., (0, 0) and with radius ‘r’ is

\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}……….(1)\]

The parametric equation of (1) is

\[x=r\cos \theta ,y=r\sin \theta \]

2. If the center of a circle is at (α, β) and radius is ‘r’ then the equation of the circle is

\[x=r\cos \theta ,y=r\sin \theta \]

3. The general equation of a circle is

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]

Whose center is at ** ( – g, – f)** and the radius is

\[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]

4. A quadratic equation in x and y

\[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\]

will represent equation of a circle if ** a = b** and

**.**

*h = 0*5. If the end point of a diameter of a circle is **(x _{1}, y_{1}) **and

**(x**then the equation of the circle is

_{2}, y_{2})\[\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0\]

6. Let there is two circles

\[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0……….(1)\]

\[{{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0……….(2)\]

Then the equation of the circle which passes through the points of intersection of (1) and (2) is

\[{{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}+k\left( {{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}} \right)=0,\left[ k\ne -1 \right]\]

7. Let us have a circle whose equation is

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0……….(1)\]

Then the equation of a circle which is concentric to the (1) is

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+k=0\]

Example 01 |

**Find the equation of the circle whose center is at (2, -3) and passing through the point (5, -1).**

**Solution:**

Let the radius of the circle be ‘r’, and since the center of the circle is (2, -3) then the equation of the circle

\[{{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}={{r}^{2}}……….(1)\]

Since (1) passes through (5, -1)

\[{{\left( 5-2 \right)}^{2}}+{{\left( -1+3 \right)}^{2}}={{r}^{2}}\]

\[\Rightarrow {{r}^{2}}=9+4=13\]

Therefore the equation of the circle is

\[{{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=13\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}-4x+6y=0\]

Example 02 |

**Find the center and radius of the circle**

\[5{{x}^{2}}+5{{y}^{2}}-8x+6y-15=0\]

**Solution:**

Representing the given equation in the general form of equation of circle, we have

\[{{x}^{2}}+{{y}^{2}}-\frac{8}{5}x+\frac{6}{5}y-3=0……….(1)\]

Again, general equation of circle is

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0……….(2)\]

Comparing (1) and (2) we get

\[2g=-\frac{8}{5},~~2f=\frac{6}{5}~~and~~c=-3\]

\[\therefore g=-\frac{4}{5},~~f=\frac{3}{5}~~and~~c=-3\]

Therefore, the coordinate of the center of (1) is

\[\left( -g,-f \right)=\left( \frac{4}{5},-\frac{3}{5} \right)\]

And radius of the circle is

\[\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{\frac{16}{25}+\frac{9}{25}-\left( -3 \right)}=\sqrt{4}=2~~unit\]

Example 03 |

**Find the parametric equation of the circle**

\[{{x}^{2}}+{{y}^{2}}-5x+2y+5=0\]

**Solution:**

\[{{x}^{2}}+{{y}^{2}}-5x+2y+5=0\]

\[\Rightarrow {{x}^{2}}-2.x.\frac{5}{2}+{{\left( \frac{5}{2} \right)}^{2}}+{{y}^{2}}+2.y.1+{{1}^{2}}={{\left( \frac{5}{2} \right)}^{2}}+1-5\]

\[\Rightarrow {{\left( x-\frac{5}{2} \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( \frac{3}{2} \right)}^{2}}……….(1)\]

\[\text{Clearly,}~~~x-\frac{5}{2}=\frac{3}{2}\cos \theta ~~and~~y+1=\frac{3}{2}\sin \theta ~~satisfies~~(1)\]

Therefore the parametric equation of the given circle is

\[x-\frac{5}{2}=\frac{3}{2}\cos \theta ~~and~~y+1=\frac{3}{2}\sin \theta ~\]

\[\Rightarrow x=\frac{1}{2}\left( 5+3\cos \theta \right)~~and~~y=-1+\frac{3}{2}\sin \theta \];