If $(a, b)$ is the midpoint of the chord $2 x-y+3=0$ of the circle $x^2+y^2+6 x-4 y+4=0$, then $2 a+3 b=$

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } 2 x-y+3=0, \Rightarrow y=2 x+3 \ & x^2+y^2+6 x-4 y+4=0 \ & \Rightarrow x^2+(2 x+3)^2+6 x-4(2 x+3)+4=0 \ & \Rightarrow 5 x^2+10 x+1=0 \end{aligned}$ Sum of roots $x_1+x_2=-\frac{10}{5}=-2$ $	herefore$ Mid point $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}ight)=\left(\frac{-2}{2}, x_1+x_2+3ight)$ $(a, b)=(-1,1) 	herefore 2 a+3 b=1$

If $(a, b)$ is the midpoint of the chord $2 x-y+3=0$ of the circle $x^2+y^2+6 x-4 y+4=0$, then $2 a+3 b=$

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