If $2 k x+3 y-1=0,2 x+y+5=0$ are conjugate lines with respect to the circle $x^2+y^2-2 x-4 y-4=0$, then $k=$

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Aicharya · Accepted Answer

Let point $\left(x_1, y_1ight)$ lie on the line $2 k x+3 y-1=0$. $x^2+y^2-2 x-4 y-4=0$ Given circle, $x^2+y^2-2 x-4 y-4=0$ Equation of chord of contact from $\left(x_1, y_1ight)$ to the circle $ \begin{aligned} & x x_1+y y_1-\frac{2\left(x+x_1ight)}{2}-\frac{4\left(y+y_1ight)}{2}-4=0 \ & \Rightarrow \quad x\left(x_1-1ight)+y\left(y_1-2ight)-x_1-2 y_1-4=0 \end{aligned} $ Now, $2 x+y+5=0$ and $2 k x+3 y+1=0$ are conjugate line $ 	herefore \quad 2 x+y+5=0 \quad 	ext { and } \quad x\left(x_1-1ight)+y\left(y_1-2ight) $ $-x_1-2 y_1-4=0$ are coincide. $ \begin{aligned} & 	herefore \quad \frac{x_1-1}{2}=\frac{y_1-2}{1}=\frac{x_1+2 y_1+4}{-5}=\lambda \ & \Rightarrow \quad x_1=2 \lambda+1, y_1=\lambda+2 \ & x_1+2 y_1+4=-5 \lambda \end{aligned} $ Put $x_1, y_1$ in $x_1-2 y_1-4=-5 k$ $ \begin{aligned} & 2 \lambda+1+2 \lambda+4+4=-5 \lambda \Rightarrow \lambda=-1 \ & 	herefore \quad x_1=-1, y_1=1 \end{aligned} $ Now, put $x_1=-1$ and $y_1=1$ $2 k x+3 y-1=0$, we get $ -2 k+3-1=0 \Rightarrow k=1 $

If $2 k x+3 y-1=0,2 x+y+5=0$ are conjugate lines with respect to the circle $x^2+y^2-2 x-4 y-4=0$, then $k=$

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