CircleHard
Question
If the circle x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is
Options
A.2 of - 3/2
B.-2 or -3/2
C.2 or 3/2
D.-2 or 3/2
Solution
Since, the given circles intersect orthogonally
∴ 2(1) (0) + 2(k) (k) = 6 + k (∵ 2g1g2 + 2f1f2 = c1 + c2)
⇒ 2k2 - k - 6 = 0
⇒
∴ 2(1) (0) + 2(k) (k) = 6 + k (∵ 2g1g2 + 2f1f2 = c1 + c2)
⇒ 2k2 - k - 6 = 0
⇒
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