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(B) The equations of the sides are:$y=0$ ... $(i)$$y=x$ ... (ii)$2x+3y=10$ ... (iii)Solving $(i)$ and (iii) gives vertex $A(5,0)$.Solving $(i)$ and (i

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$\left(\frac{5}{2},-\frac{1}{2}\right)$

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(B) The equations of the sides are:$y=0$ ... $(i)$$y=x$ ... (ii)$2x+3y=10$ ... (iii)Solving $(i)$ and (iii) gives vertex $A(5,0)$.Solving $(i)$ and (ii) gives vertex $B(0,0)$.Solving (ii) and (iii) gives vertex $C(2,2)$.Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... (iv).Since it passes through $B(0,0)$,$c=0$.Since it passes through $A(5,0)$,$25+10g=0 \Rightarrow g=-5/2$.Since it passes through $C(2,2)$,$4+4+4g+4f=0 \Rightarrow g+f+2=0$.Substituting $g=-5/2$,we get $-5/2+f+2=0 \Rightarrow f=1/2$.The centre of the circle is $(-g, -f) = (5/2, -1/2)$.

The centre of the circle which passes through the vertices of the triangle formed by the lines $y=0$,$y=x$ and $2x+3y=10$ is

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