Question
If the circles $${x^2} + {y^2} + 2x + 2ky + 6 = 0,\,\,{x^2} + {y^2} + 2ky + k = 0$$ intersect orthogonally, then $$k$$ is-
A.
$$2\,\,{\text{or }} - \frac{3}{2}$$
B.
$$ - 2\,\,{\text{or }} - \frac{3}{2}$$
C.
$$2\,\,{\text{or }}\frac{3}{2}$$
D.
$$ - 2\,\,{\text{or }}\frac{3}{2}$$
Answer :
$$2\,\,{\text{or }} - \frac{3}{2}$$
Solution :
$$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$$ (formula for orthogonal intersection of two cricles)
$$\eqalign{
& \Rightarrow 2(1)(0) + 2(k)(k) = 6 + k \cr
& \Rightarrow 2{k^2} - k - 6 = 0 \cr
& \Rightarrow k = - \frac{3}{2},\,\,2\, \cr} $$