If the circles `x^2+y^2+2x+2ky+6=0` and `x^2+y^2+2ky+k=0` intersect orthogonally then k equals (A) `2 or -3/2` (B) `-2 or -3/2` (C) `2 or 3/2` (D) `-2 or 3/2`A. 2 or -3/2B. `-2 or -3//2`C. 2 or 3/2D. `-2 or 3//2`

If the circles `x^2+y^2+2x+2ky+6=0` and `x^2+y^2+2ky+k=0` intersect or

1.	If the circles `x^2+y^2+2x+2ky+6=0` and `x^2+y^2+2ky+k=0` intersect orthogonally then k equals (A) `2 or -3/2` (B) `-2 or -3/2` (C) `2 or 3/2` (D) `-2 or 3/2`A. 2 or -3/2B. `-2 or -3//2`C. 2 or 3/2D. `-2 or 3//2`
Answer» Correct Answer - A Since, the given circles intersect orthogonally. `therefore 2(1) (0)+2(k)(k)=6+k` `[therefore2g_(1)+2f_(1)f_(2)=c_(1)+c_(2)]` `rArr 2k^(2)-k-6=0rArrk=-(3)/(2),2`

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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 equals (A) `2 or -3/2` (B) `-2 or -3/2` (C) `2 or 3/2` (D) `-2 or 3/2`A. 2 or -3/2B. `-2 or -3//2`C. 2 or 3/2D. `-2 or 3//2`

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