Before deriving the equation of a circle, let us focus on what is a circle?. A circle is a set of all points which are equally spaced from a fixed point in a plane. The fixed point is called the centre of the circle. The distance between the centre and any point on the circumference is called the radius of the circle. In this article, we are going to discuss what is an equation of a circle formula in standard form, and finding the equation of a circle when the centre is origin and centre is not an origin with examples.

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## What is the Equation of a Circle?

A circle is formed when an arc is drawn from the fixed point called the centre, in which all the points on the curve are having the same distance from the centre point of the centre. We know that there is a question arises in case of circle whether being a function or not. It is clear that a circle is not a function. Because, a function is defined by each value in the domain is exactly associated with one point in the codomain, but a line that passes through the circle, intersect the line at two points on the surface. The mathematical way to describe the circle is an equation. Here, the equation of the circle is provided in all the forms such as general form, standard form along with the examples.

## Equation of a Circle When the Centre is Origin

Consider an arbitrary pointÂ P(x,y)Â on the circle. LetÂ aÂ be the radius of the circle which is equal toÂ OP.

We know that the distance between the pointÂ (x,y)Â and originÂ (0,0)can be found using the distance formula which is equal to-

âˆš[x^{2}+Â y^{2}]= a

Therefore, the equation of a circle, with the centre as the origin is,

**x ^{2}+y^{2}= a^{2}**

Where “a” is the radius of the circle.

## Equation of a Circle When the Centre is not an Origin

LetÂ C(h,k)Â be the centre of the circle andÂ P(x,y)Â be any point on the circle.

Therefore the radius of a circle is CP.

By using distance formula,

(x-h)^{2Â }+ (y-k)^{2} = CP^{2}

Let radius be a.

Therefore, the equation of the circle with centre (h,k)and the radius aÂ is,

**(x-h) ^{2}+(y-k)^{2} = a^{2}**

which is called the standard form for the equation of a circle.

### Equation of a Circle in General Form

The general equation of any type of circle is represented by:

x^{2}Â +Â y^{2}Â +Â 2gxÂ +Â 2fyÂ +Â cÂ =Â 0, for all values ofÂ g,Â fÂ andÂ c.

AddingÂ g^{2}Â +Â f^{2}Â on both sides of the equation gives,

x^{2}Â +Â 2gxÂ +Â g^{2}+Â y^{2}Â +Â 2fyÂ +Â f^{2}=Â g^{2}Â +Â f^{2}Â âˆ’Â cÂ â€¦â€¦â€¦â€¦â€¦â€¦(1)

Since,Â (x+g)^{2Â }=Â x^{2}+Â 2gxÂ +Â g^{2}Â andÂ (y+f)^{2Â }=y^{2Â }+Â 2fyÂ +Â f^{2}Â substituting the values in equation (1), we have

(x+g)^{2}+Â (y+f)^{2Â }=Â g^{2Â }+Â f^{2}âˆ’cÂ â€¦â€¦â€¦â€¦â€¦.(2)

Comparing (2) withÂ (xâˆ’h)^{2Â }+Â (yâˆ’k)^{2Â }=Â a^{2}, whereÂ (h,k)Â is the centre andÂ aÂ is the radius of the circle.

h=âˆ’g,Â k=âˆ’f

aÂ =Â g^{2}+Â f^{2}âˆ’c

Therefore,

x^{2Â }+Â y^{2Â }+Â 2gxÂ +Â 2fyÂ +Â cÂ =Â 0, represents the circle with centreÂ (âˆ’g,âˆ’f)Â and radius equal toÂ aÂ =Â g^{2Â }+Â f^{2}âˆ’Â c.

- IfÂ g
^{2Â }+Â f^{2Â }>Â c, then the radius of the circle is real. - IfÂ g
^{2Â }+Â f^{2Â }=Â c, then the radius of the circle is zero which tells us that the circle is a point which coincides with the centre. Such type of circle is called a point circle - g
^{2Â }+Â f^{2Â }<c, then the radius of the circle become imaginary. Therefore it is a circle having a real centre and imaginary radius.

### How to Find the Equation of the Circle?

Here, some solved problems are given to find the equation of a circle on both cases such as when the centre of a circle is origin and centre is not an origin is given below.

**Example 1:**

Consider a circle whose centre is at the origin and radius is equal to 8 units.

**Solution:**

Given: Centre is (0, 0), radius is 8 units.

We know that the equation of a circle when the centre is origin:

**x ^{2}+Â y^{2Â }= a^{2}**

For the given condition, the equation of a circle is given as

x^{2Â }+Â y^{2Â }= 8^{2}

x^{2Â }+Â y^{2}= 64, which is the equation of a circle

**Example 2:**

Find the equation of the circle whose centre is (3,5)Â and the radius is 4 units.

**Solution:**

Here, the centre of the circle is not an origin.

Therefore, the general equation of the circle is,

(x-3)^{2Â }+ (y-5)^{2} Â = 4^{2}

x^{2Â }– 6x + 9 + y^{2Â }-10y +25Â = 16

x^{2Â }+y^{2Â }-6x -10y + 18Â =0

**Example 3:**

Equation of a circle isÂ x^{2}+y^{2}âˆ’12xâˆ’16y+19=0. Find the centre and radius of the circle.

**Solution:**

Given equation is of the formÂ x^{2}+Â y^{2}Â +Â 2gxÂ +Â 2fyÂ +Â cÂ =Â 0,

2gÂ =Â âˆ’12,Â 2fÂ =Â âˆ’16,cÂ =Â 19

gÂ =Â âˆ’6,fÂ =Â âˆ’8

Centre of the circle isÂ (6,8)

Radius of the circle =Â âˆš[(âˆ’6)^{2}Â +Â (âˆ’8)^{2}Â âˆ’Â 19Â ]=Â âˆš[100Â âˆ’Â 19]Â =

=Â Â âˆš81 =Â 9Â units.

Therefore, the radius of the circle isÂ 9 units.

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