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

Question

Suppose that two chords, drawn from the point $(1,2)$ on the circle $x^2+y^2+x-3 y=0$ are bisected by the $y$-axis. If the other ends of these chords are R and S , and the mid point of the line segment RS is $(\alpha, \beta)$

Accepted Answer

The given circle is

${x}^{2}+{y}^{2}+x-3y=0$

and the point from which the two chords are drawn is

$P(1,2)$

We are told that these chords are bisected by the $y$-axis.

Let the other end of one such chord be $R(x,y)$.

Since the chord joining $P(1,2)$ and $R(x,y)$ is bisected by the $y$-axis, its midpoint must lie on the $y$-axis.

So the $x$-coordinate of the midpoint is $0$.

Midpoint of $P(1,2)$ and $R(x,y)$ is

$(\frac{1+x}{2},\frac{2+y}{2})$

Since this lies on the $y$-axis,

$\frac{1+x}{2}=0$

So,

$x=-1$

Thus, the other end of each such chord must lie on the line

$x=-1$

Now find where this line cuts the circle.

Substitute $x=-1$ in

${x}^{2}+{y}^{2}+x-3y=0$

We get

$(-1{)}^{2}+{y}^{2}+(-1)-3y=0$

$1+{y}^{2}-1-3y=0$

${y}^{2}-3y=0$

$y(y-3)=0$

So,

$y=0\, \, y=3$

Hence the two points are

$R(-1,0),\, S(-1,3)$

Now midpoint of $RS$ is

$(\alpha ,\beta )=(\frac{-1+(-1)}{2},\frac{0+3}{2})$

$(\alpha ,\beta )=(-1,\frac{3}{2})$

Therefore,

$\alpha +\beta =-1+\frac{3}{2}=\frac{1}{2}$

So,

$6(\alpha +\beta )=6\cdot \frac{1}{2}=3$

Hence the correct answer is

$3$

So, the correct option is Option B.

Solutions · 1