The equation of the circle having centre $(1,\ -2)$ ;and passing through the point of intersection of the lines $3x+y=14$ and $2x+5y=18$ is

The circle passes through intersection of $3x+y=14$ and $2x+5y=18$.

$\therefore 5(3x+y-14)-2x-5y+18=0\implies 15x-70-2x+18=0\implies x=4\implies y=2$

Hence, equation of circle is

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

$\therefore x^2-2x+1+y^2+4y+4=25$

$\therefore x^2+y^2-2x+4y-20=0$

This is the required equation.