A Circle is defined as the locus of all points which are equidistance from a point. The point is called the Centre of the circle. So, the Centre of a Circle is a point which lies inside the circle and is at equidistance from all the points on the circumference of the circle. A circle is named by the centre of a Circle.

The constant distance from the centre of the circle to any point on the circumference of the circle is called the radius of the circle. The Diameter of the circle is the line segment that joins any two points on a circle and that passes through the centre of the Circle.
Center of a Circle
In the circle on the left, “A” is the centre of the circle. B, C, D are three points on the circumference of the circle which are at equidistance from A. Thus AB, AC, AD are all the radii.

CD becomes the diameter of the Circle. The diameter of the circle is twice the radius of the circle. So, CD = 2r.

Real World Examples: Few real world examples of a circle are a wheel in which wheel hub is the centre, a pizza is cut from its centre, flowers with pollen.

In general, if an equation of a circle is given by x$^2$+ y$^2$+ a x + b y = c then,
The centre is given by the coordinates (-$\frac{a}{2}$, -$\frac{b}{2}$) and the radius, r such that r$^2$ = c + ($\frac{a}{2}$)$^2$ + ($\frac{b}{2}$)$^2$Otherwise, if an equation of the circle is given in its centre-radius form, (x - h)$^2$ + (y - k)$^2$ = r$^2$, then

The centre is given by (h, k) and radius r.
Case (i): If the circle equation is given in its centre-radius form, (x - h)$^2$ + (y - k)$^2$ = r$^2$
Then the centre, C = (h, k).

Case (ii): If the circle equation is given by x$^2$+ y$^2$+ ax + by=c then,
  • Take the coefficients of x and y terms and write them as ordered pair (a, b)
  • Divide by 2 as ($\frac{a}{2}$, $\frac{b}{2}$).
  • Change the sign. (-$\frac{a}{2}$, -$\frac{b}{2}$).
  • The obtained order pair is the centre.

Case (iii): If the circle equation is given in general form as ax$^2$ + by$^2$ + cx + dy + e = 0. To find the centre of the circle from this equation we use the technique of completing the square to turn the quadratic into the sum of a squared binomial and a number. This would reduce the general form to the centre-radius form of the circle equation which is (x - h)$^2$ + (y - k)$^2$ =r$^2$. In this the centre, C = (h, k)

The following steps helps in reducing the general equation to centre-radius form:

1. Eliminate the constant “e” on the left side by performing reverse operation on it. (As if moving “e” to the other side). Which means reducing the equation from ax$^2$+by$^2$+cx + dy + e = 0 to ax$^2$ + by$^2$ + cx+dy = -e.

2. Group the x terms together and y terms together. (ax$^2$ + cx)+(by$^2$ + dy) = -e.

3. Divide the equation by “a” (which means whatever is multiplied on the squared terms; divide it off from every term). (x$^2$ + $\frac{c}{a}$ x)+($\frac{b}{a}$ y$^2$ + $\frac{d}{a}$ y)= -$\frac{e}{a}$

4. Now we have a x$^2$ term with x term grouped and y$^2$ term with y term grouped. This is where the squaring term is added. Take the x-term coefficient, multiply it by $\frac{1}{2}$, square it, and then add this to both sides of the equation. Do the same with the y-term coefficient also. Change the left side to squared form, and simplify on the right side.
(x$^2$ + $\frac{c}{a}$ x+ ($\frac{c}{2a}$)$^2$ )+($\frac{b}{a}$ y$^2$ + $\frac{d}{a}$ y + ($\frac{d}{2a}$)$^2$) = -$\frac{e}{a}$ + ($\frac{c}{2a}$)$^2$ + ($\frac{d}{2a}$)$^2$

5. This can be reduced to the form which is (x - h)$^2$ + (y - k)$^2$ = r$^2$.

6. The centre is at (h, k)

Solved Examples

Question 1: Find the centre of the circle with equation x$^2$ + y$^2$ - 6x - 4y + 9 = 0
Solution:
 
The given equation is in the form of x$^2$ + y$^2$ + ax + by = c. So we will find the centre as described in case (ii).

    Take the coefficients of x and y terms and write them as ordered pair (a, b). Here a = -6 and b =-4. So the ordered pair would be (-6, -4)

    Divide by 2 as ($\frac{a}{2}$, $\frac{b}{2}$). i.e., (-3, -2)

    Change the sign.   (-$\frac{a}{2}$, -$\frac{b}{2}$). i.e., (3, 2)

    The obtained order pair is the centre. Thus the centre = (3, 2).
 

Question 2: Find the centre of the circle with equation  4x$^2$ + 4y$^2$ - 16x - 8y + 16 = 0.
Solution:
 
To find the centre of the circle, we first rewrite the given equation into the centre- radius form or the standard form which is (x - h)$^2$ + (y - k)$^2$ = r$^2$. Let us follow the steps mentioned above.

Eliminate the constant on the left side by performing reverse operation on it.
4x$^2$ + 4y$^2$ - 16x - 8y = -16  

Group the x terms together and y terms together. (4x$^2$ - 16x) + (4 y$^2$ - 8y) = -16.

Divide the equation by “4”. ($\frac{4}{4}$ x$^2$ - $\frac{16}{4}$ x) + ($\frac{4}{4}$ y$^2$ - $\frac{8}{4}$ y) = -$\frac{16}{4}$. This gives (x$^2$ - 4 x) + (y$^2$ - 2y) = -4.

Now we have  x$^2$  term with x term grouped and y$^2$ term with y term grouped. This is where the squaring term is added. Take the x-term coefficient which is -4, multiply it by $\frac{1}{2}$ thus we get -2, squaring it we get 4, and then add this to both sides of the equation. Do the same with the y-term coefficient also. This means we add 1 to both sides of the equation. Change the left side to squared form, and simplify on the right side.

(x$^2$ - 4x + (2)$^2$) + (y$^2$ - 2y + (1)$^2$) = -4 + (2)$^2$ +(1)$^2$

(x$^2$ - 2 $\times$ 2 $\times$ x+(2)$^2$) + (y$^2$ - 2 $\times$ 1 $\times$ y+1$^2$) = -4 + (2)$^2$ + (1)$^2$

This can be reduced to the form which is (x - h)$^2$ + (y - k)$^2$ = r$^2$.
(x -2)$^2$ + (y-1)$^2$ = 1$^2$.

The centre is at (h, k) = (2, 1)

Thus, the centre of the circle with equation  4x$^2$ +4y$^2$ - 16x - 8y + 16 = 0 is (2, 1).