If lines 2x−3y+6=0 and kx+2y+12=0 cut the coordinate axes in the concylic points, then the value of k is
2x−3y+6=0 meets the axes at A−3,0 and B0,2. kx+2y+12=0 meets the axes at the D0,−6 and C−12/k,0.OA×OC=OB×OD
⇒3×12/k=2×6
⇒k=−3