The new coordinates after shift are given by, x′=x−h y′=y−k So x2+5xy+2y2+5x+6y+7=0 (x′+h)2+5(x′+h)(y′+k)+2(y′+k)2+5(x′+h)+6(y′+k)+7=0 x′2+2hx′+h2+5xy′+5kx′+5y′h+5hk+2y′2+2k2+4y′k+5x′+5h+6y′+6k+7=0 The equation is free from the first order so 2h+5k+5=0⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) 5h+4k+6=0⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) From equation (I) and (II), h=−