(−k,−2)→ Centre of circle x2+y2+2kx+4y−4=0 since it lies in fourth quadrant −k>0⇒k<0 Centre of circle x2+y2+6x−2y+6=0 is (−3,1) and radius =√9+1−6 ∵r1+r2=c1c2 √k2+4+4+√9+1−6=√(k−3)2+(−2−1)2 √k2+8+2=√k2−6k+18 k=−1 satisfies the equation.