Points (6,−k) and (−3,2) are conjugate points to the circle x2+y2+6x+4y+12=0 ∴ We know that if (x1,y1) and (x2,y2) are conjugate points of circle x2+y2+2gx+2fy+c=0 Then, x1x2+y1y2+g(x1+x2)+f(y1+y2)+c=0 ∴ (6) (−3)+(−k)(2)+3(6−3)+2(−k+2)+12=0 ⇒−18−2k+9−2k+4+12=0 ⇒4k=7 ⇒k=