Question:
If the circles x2 + y2 + 2x + 2ky + 6 = 0, x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is
If the circles x2 + y2 + 2x + 2ky + 6 = 0, x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is
Updated On: Oct 4, 2024
- 2(or)\(\frac{-3}{2}\)
- -2(or)\(\frac{-3}{2}\)
- 2(or)\(\frac{3}{2}\)
- -2(or)\(\frac{3}{2}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): 2(or)\(\frac{-3}{2}\)
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