If the circles x2+y2+2x+2ky+6=0 and x2+y2+2ky+k=0 intersect orthogonally, then k is :

(A) 2 or –3/2 (B) –2 or 3/2
(C) 2 or 3/2 (D) (2,+∞)

R2 + r2 = distance between centers is condition of orthogonality

the circles are : (x+1)²+(y+k)² = k²-5

                       x² + (y+k)² = k²-k

its : k²-5+k²-k =  1

      which gives k = 2 , -3/2

option A

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IIT Kgp - 05 batch