Question

If the circles x2 + y2 + 2x + 2ky + 6 = 0, x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is

• A. 2(or)$\frac{-3}{2}$
• B. -2(or)$\frac{-3}{2}$
• C. 2(or)$\frac{3}{2}$
• D. -2(or)$\frac{3}{2}$

Answer 1

Answer: (A)

2(or)$\frac{-3}{2}$

Answer 2

The correct option is (A): 2(or)$\frac{-3}{2}$