A circle passes through $(0,0),(a, 0)$ and $(0, b)$ the co-ordinate of its centre are
A) $\dfrac{a}{2}, \dfrac{b}{2}$
B) $\dfrac{b}{2}, \dfrac{a}{2}$
C) $(a, b)$
D) $(b, a)$

Hint: We know that the centre of a circle is a location inside the circle that is situated in the middle of the circumference. We have to find the coordinates of its centre. For that, we use the general equation of the circle. The radius of a circle is the constant distance from the circle's centre to any point on the circle.

Complete step by step Solution:
The equation of the circle for each of the given points can be represented as
$(0,0) \rightarrow c=0$
$(a,0)\to {{a}^{2}}+2ga+c$
$=0\to a(a+2g)$
$-\dfrac{a}{2}$
For the point
$(0,b)\to b(b+2f)=0$
$(0, b) \rightarrow b(b+2 f)=0 \Rightarrow f=-\dfrac{b}{2}$
Center of circle $=(-\mathrm{g},-\mathrm{f})=\left(\dfrac{\mathrm{a}}{2}, \dfrac{\mathrm{b}}{2}\right)$
The general equation of a circle is another name for the centre of a circle formula. If the radius is r, the centre's coordinates are $(h,k),$and any point on the circle is$(x, y)$, the centre of the circle formula is as follows:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
The equation for the centre of the circle is another name for this. In the sections that follow, we'll use this formula to determine a circle's equation or centre.

Therefore, the correct option is (A).


Note: Note that the group of points whose separation from a fixed point has a constant value is represented by a circle. The radius of the circle abbreviated $r$, is a constant that describes this fixed point, which is known as the circle's centre. The formula for a circle$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$