When the given lines are conjugate with the circle, then (g2+f2−c)(a1a2+b1b2)=(a1g+b1f−c1)(a2g+b2f−c2) Substitute the values [(−1)2+(−1)2−(−1)][1×1+k×1]=[1×(−1)+k×(−1)−(−4)]×[1×(−1)+1×(−1)]−(−5) 3(k+1)=3(3−k) k=1