The circles $x^2 + y^2 + 2x + 2ky + 6 = 0$ and $x^2 + y^2 + 2ky + k = 0$ intersect orthogonally. The value(s) of $k$ is (are):
(1) $2$ or $-\frac{3}{2}$
(2) $-2$ or $\frac{3}{2}$
(3) $2$ or $\frac{3}{2}$
(4) $-2$ or $-\frac{3}{2}$
Step-by-Step Solution
Key Concept: Apply the orthogonality condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Here $(g_1, f_1, c_1) = (1, k, 6)$ and $(g_2, f_2, c_2) = (0, k, k)$, giving $2k^2 - k - 6 = 0$, i.e., $(2k + 3)(k - 2) = 0$.
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Correct Answer: (1)