Given equation of sides are
$y=0\,\,\,\,\,\,\dots(i)$
$y=x \,\,\,\,\,\,\dots(ii)$
$2 x+3 y=10 \,\,\,\,\,\,\dots(iii)$
Solving Eqs. (i) and (iii) to get vertex 'A'
$\therefore \, A=(5,0)$
Solving (i) and (ii) to get vertex 'B'
$B=(0,0)$
Solving Eqs. (ii) and (iii) to get vertex 'C'
$C=(2,2)$
Let equation of circle be $x^{2}+y^{2}+2 g x+2 f y+c=0\,\,\,\,\,\ldots (iv) $
Since, Eq. (iv) passes through $B(0,0)$
$\Rightarrow \, C=0$
Since, Eq. (iv) passes through $A(5,0)$
$25+0+10 g+0+0=0 $
$g=-5 / 2$
Eq. (iv) passes through $ C(2,2)$
$\therefore \, 4+4+4 g+4 y''=0 $
$g+f+2 =0 $
$-\frac{5}{2}+f+2 =0$
$f =\frac{1}{2}$
$\therefore $ Centre of circle $=(-g,-f)=\left(\frac{5}{2}, \frac{-1}{2}\right)$
