If the line y=mx+1 meets the circle x^2+y^2+3x=0 in two points equidistant from and on opp…

Question

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

Accepted Answer

Circle: centre $\left(-\tfrac32,0\right)$, radius $\tfrac32$. The line meets the $y$-axis at $A(0,1)$. Since the two intersection points are symmetric about the $x$-axis, the chord must be perpendicular to it in a way that makes the line's inclination satisfy $\tan\theta=\dfrac{OA}{OB}$ where $B=\left(-\tfrac32,0\right)$ is the centre:
$$m=\frac{1}{3/2}=\frac23\ \Rightarrow\ 3m-2=0.$$

Solutions · 1