If $x+y-1=0$ and $2 x-y+1=0$ are conjugate lines with respect to a circle $x^2+y^2-4 x+2 f y-1=0$, then $f=$

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Aicharya · Accepted Answer

Conjugate lines are : $x+y-1=0,2 x-y+1=0$ $\mathrm{l}_1=1, \mathrm{~m}_1=1, \mathrm{n}_1=-1, \mathrm{l}_2=2, \mathrm{~m}_2=-1, \mathrm{n}_2=1$ $\begin{aligned} & 	herefore \quad g=-2, c=-1 \ & 	ext { Radius, } R=\sqrt{f^2+4+1}=\sqrt{f^2+5}\end{aligned}$ Now, the condition for the conjugate lines : $\begin{aligned} & \mathrm{R}^2\left(\mathrm{l}_1 \mathrm{l}_2+\mathrm{m}_1 \mathrm{~m}_2ight)=\left(\mathrm{l}_1 \mathrm{~g}+\mathrm{m}_1 \mathrm{f}-\mathrm{n}_1ight)\left(\mathrm{l}_2 \mathrm{~g}+\mathrm{m}_2 \mathrm{f}-\mathrm{n}_2ight) \ & \left(\mathrm{f}^2+5ight)(1.2+1(-1))=(1(-2)+1 \mathrm{f}+1)(2(-2)+(-1) \ & \mathrm{f}-1) \ & \Rightarrow \mathrm{f}^2+5=(\mathrm{f}-1)(-\mathrm{f}-5) \ & \Rightarrow \mathrm{f}^2+2 \mathrm{f}=0 \ & \Rightarrow \mathrm{f}=0 	ext { or }-2 .\end{aligned}$

If $x+y-1=0$ and $2 x-y+1=0$ are conjugate lines with respect to a circle $x^2+y^2-4 x+2 f y-1=0$, then $f=$

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