If the circles \(x^{2}+y^{2}+2 x+2 k y+6=0\) and \(x^{2}+y^{2}+2 k y+k=0\) intersect orthogonally, then \(k\) is
Step-by-step Solution:
Radius of first circle is \(\sqrt{1^{2}+k^{2}-6}=\sqrt{k^{2}-5}\) Radius of second circle \(=\sqrt{k^{2}-k}\) \[\] Distance between their centres \[\] \(=\sqrt{(-1-0)^{2}+(k-k)^{2}}=1\) \[\] Circles are cutting orthogonally if \[\] \(\left(k^{2}-5\right)+\left(k^{2}-k\right)=1 \Rightarrow 2 k^{2}-k-6=0\) \[\] \(\Rightarrow(2 k+3)(k-2)=0\) or \(k=-\frac{3}{2}\) or 2 \[\] Choice (A)\[\]