If the line $y=m x+1$ meets the circle $x^2+y^2+3 x=0$ in two points equidistant from and on opposite sides of $x$-axis, then

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Aicharya · Accepted Answer

Circle: $x^2+y^2+3 x=0$ Centre, $B=\left(-\frac{3}{2}, 0\right)$

$y$-intercept of the line $=1$ $ \therefore \quad \mathrm{A}=(0,1) $ Slope of line, $m=\tan \theta=\frac{O A}{O B}$ $ \begin{aligned} & \Rightarrow m=\frac{1}{\frac{3}{2}}=\frac{2}{3} \\ & \Rightarrow 3 m-2=0 \end{aligned} $

If the line $y=m x+1$ meets the circle $x^2+y^2+3 x=0$ in two points equidistant from and on opposite sides…

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