# Equation Of A Circle

Before deriving the equation of a circle, let us focus on 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 centre and any point on the circumference is called the radius of the circle.

## Equation of a Circle:

### Centre is Origin:

Consider an arbitrary point $P(x,y)$ on the circle. Let $a$ be the radius of the circle which is equal to $OP$.

We know that, distance between the point $(x,y)$ and origin $(0,0)$can be found using the distance formula which is equal to-

$\sqrt{x^{2}+y^{2}} = a$

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

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

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

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

$x^{2}+y^{2} = 8^{2}$

### Centre is not origin:

Let $C(h,k)$ be the centre of the circle and $P(x,y)$ be any point on the circle.

Therefore radius of a circle is CP.

By using distance formula,

$(x-h)^{2}+(y-k)^{2} = CP^{2}$

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

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

 Example: Find the equation of the circle whose centre is $(3,5)$ and radius is 4 units. Solution: Here, centre of the circle is not origin. Therefore, the general equation of the circle is, $(x-3)^2+(y-5)^2$ = $4^2$ $x^2-6x+9+y^2-10y+25$ = $16$ $x^2+y^2-6x-10y+18$ = $0$

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

$x^2~+~y^2~+~2gx~+~2fy~+~c$ = $0$, for all values of $g$,$f$ and $c$.

Adding $g^2~+~f^2$ on both sides of the equation gives,

$x^2~+~2gx~+~g^2~+~y^2~+~2fy~+~f^2$ = $g^2~+~f^2~-~c$ ………………(1)

Since, $(x+g)^2 = x^2+2gx+g^2$ and $(y+f)^2 = y^2+2fy+f^2$ substituting the values in equation (1), we have

$(x+g)^2+(y+f)^2 = g^2+f^2-c$ …………….(2)

Comparing (2) with $(x-h)^2+(y-k)^2 = a^2$, where $(h,k)$ is the centre and $a$ is the radius of the circle.

$h =-g$, $k = -f$

$a = g^{2}+ f^{2} -c$

Therefore,

$x^2+y^2+2gx+2fy+c$ = $0$, represents the circle with centre $(-g,-f)$ and radius equal to $a = g^{2}+ f^{2} -c$.

• If $g^2+f^2>c$, then the radius of the circle is real.
• If $g^2+f^2 = c$, 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 as point circle
• $g^2+f^2<c$, then the radius of the circle become imaginary. Therefore it is a circle having real centre and imaginary radius.
 Example: Equation of a circle is $x^2+y^2-12x-16y+19 = 0$. Find centre and radius of the circle. Solution: Given equation is of the form $x^2~+~y^2~+~2gx~+~2fy~+~c$ = $0$, $2g$ = $-12$, $2f$ = $-16$,$c$ = $19$ $g$ = $-6$,$f$ = $-8$ Centre of the circle is $(6,8)$ Radius of the circle = $√{(-6)^2~+~(-8)^2~-~19}$ = $√{100~-~19}$ = $9$ units.

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#### Practise This Question

The equation of circle passing through (4, 5) and having the centre at (2, 2), is