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

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

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