Question
Suppose the two chords, drawn from the point ( 1 , 2) on the circle $x^{2} + y^{2} + x - 3y = 0$ 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 $(\alpha,\beta)$, then $6(\alpha + \beta)$ is equal to :
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Solution
Sol. [Key Concept: A chord's midpoint on the y-axis fixes the far end by reflection; impose the circle equation.]
chord PS bisected by $y -$ axis
at point $M(0,\lambda)$
$$\therefore S( - 1,2\lambda - 2) $$Which satisfy equation of circle
$${\Rightarrow ( - 1)^{2} + (2\lambda - 2)^{2} - 1 - 3(2\lambda - 2) = 0 }{\Rightarrow \lambda = 1\ \&\ \lambda = \frac{5}{2} }$$Therefore two such chord PS & PR will exist
When $\lambda = 1 \Rightarrow \text{ }S( - 1,0)$
When $\lambda = \frac{5}{2}\ R( - 1,3)$
∴ Mid point of $R$ & $S$ will be $\left( - 1,\frac{3}{2} \right)$
$${\therefore\alpha = - 1,\beta = \frac{3}{2} }{\therefore 6(\alpha + \beta) = 3 }$$
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