The coordinates of the circumcentre of the triangle

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be O(0,0), A(a,0) and B(0,b).

Let the circumcentre of the triangle be represented by the point R(x, y).

So we have $O R=A R=B R$

$\mathrm{OR}=\sqrt{(-x)^{2}+(-y)^{2}}$

$\mathrm{AR}=\sqrt{(a-x)^{2}+(-y)^{2}}$

$\mathrm{BR}=\sqrt{(-x)^{2}+(b-y)^{2}}$

Equating the first pair of these equations we have,

$O R=A R$

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

Squaring on both sides of the equation we have,

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

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

$2 a x=a^{2}$

$x=\frac{a}{2}$

Equating another pair of the equations we have,

$\mathrm{OR}=\mathrm{BR}$

$\sqrt{(-x)^{2}+(-y)^{2}}=\sqrt{(-x)^{2}+(b-y)^{2}}$

Squaring on both sides of the equation we have,

$(-x)^{2}+(-y)^{2}=(-x)^{2}+(b-y)^{2}$

$x^{2}+y^{2}=x^{2}+b^{2}+y^{2}-2 b y$

$2 b y=b^{2}$

$y=\frac{b}{2}$

Hence the correct choice is option (b).