# Equation of a circle – Geometry

Geometry / Monday, November 4th, 2019

# 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 (x1, y1) and (x2, y2) then the equation of the circle is

$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$

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

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

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