Suppose two chords are drawn from the point P(1,2) on the circle x^2+y^2+x-3y=0, and each …

Question

Suppose two chords are drawn from the point $P(1,2)$ on the circle $x^2+y^2+x-3y=0$, and each is bisected by the $y$-axis. If the other ends of these chords are $R$ and $S$, and the midpoint of $RS$ is $(\alpha,\beta)$, then $6(\alpha+\beta)$ is equal to:

Accepted Answer

If the other end is $(x,y)$, the midpoint of $P(1,2)$ and $(x,y)$ lies on the $y$-axis: $\dfrac{1+x}{2}=0\Rightarrow x=-1$.

Substituting $x=-1$ in the circle: $1+y^2-1-3y=0 \Rightarrow y^2-3y=0 \Rightarrow y=0,3$.

So $R=(-1,0),\ S=(-1,3)$, midpoint $(\alpha,\beta)=\left(-1,\dfrac32\right)$.

$$6(\alpha+\beta)=6\left(-1+\frac32\right)=3.$$

Solutions · 1