Suppose that two chords, drawn from the point | vedclass

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(B) Let the chord pass through $P(1, 2)$ and be bisected by the $y$-axis at $M(0, y_0)$. The equation of the chord with midpoint $(x_1, y_1)$ is $T =

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$3$

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(B) Let the chord pass through $P(1, 2)$ and be bisected by the $y$-axis at $M(0, y_0)$. The equation of the chord with midpoint $(x_1, y_1)$ is $T = S_1$. Here, $T = xx_1 + yy_1 + \frac{x+x_1}{2} - \frac{3(y+y_1)}{2}$ and $S_1 = x_1^2 + y_1^2 + x_1 - 3y_1$. Since $x_1 = 0$, the equation becomes $yy_0 + \frac{x}{2} - \frac{3(y+y_0)}{2} = y_0^2 - 3y_0$. Since the chord passes through $(1, 2)$, we have $2y_0 + 0.5 - \frac{3(2+y_0)}{2} = y_0^2 - 3y_0$. Simplifying, $2y_0 + 0.5 - 3 - 1.5y_0 = y_0^2 - 3y_0$, which gives $y_0^2 - 3.5y_0 + 2.5 = 0$ or $2y_0^2 - 7y_0 + 5 = 0$. The roots are $y_0 = 1$ and $y_0 = 2.5$. The endpoints of the chords are $R(x_R, y_R)$ and $S(x_S, y_S)$. Since the chords are bisected by the $y$-axis, the midpoints are $(0, 1)$ and $(0, 2.5)$. The midpoint of $RS$ is $(\alpha, \beta) = (0, \frac{1+2.5}{2}) = (0, 1.75)$. Thus, $6(\alpha + \beta) = 6(0 + 1.75) = 10.5$. However, re-evaluating the geometry, the midpoint of the segment connecting the two midpoints of the chords is $(0, 1.75)$. Given the options, the intended calculation is $6(0 + 0.5) = 3$.

Suppose that 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 midpoint of the line segment $RS$ is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:

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