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If the circles x 2 +y 2 +2x+2ky+6=0 and x 2 +y 2 +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,+∞)

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,+∞)

Grade:12

1 Answers

Ramesh V
70 Points
12 years ago

4143-285_5850_img1.png

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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Ramesh  
IIT Kgp - 05 batch



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