Explanation:
C Two circles \(x^2+y^2+2 g_1 x+2 f_1 y+c_1=0\) and \(\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{~g}_2 \mathrm{x}+2 \mathrm{f}_2 \mathrm{y}+\mathrm{c}_2=0\) cuts orthogonally if \(2 \mathrm{~g}_1 \mathrm{~g}_2+2 \mathrm{f}_1 \mathrm{f}_2=\mathrm{c}_1+\mathrm{c}_2\)
Given equations of two circles are
\(x^2+y^2+2 \lambda x+6 y+1=0\)
\(x^2+y^2+4 x+2 y=0\)
On comparing (i) and (ii) with original equation, we get
\(\mathrm{g}_1=\lambda, \mathrm{f}_1=3, \mathrm{c}_1=1 \text { and } \mathrm{g}_2=2, \mathrm{f}_2=1, \mathrm{c}_2=0\)
So, from orthogonality condition, we have
\(4 \lambda+6=1 \Rightarrow 4 \lambda=-5\)
\(\therefore \lambda=\frac{-5}{4}\)