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```				   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,+∞)
```

7 years ago

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```
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
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,Ramesh   IIT Kgp - 05 batch
```
7 years ago

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