Question

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:

• A. 1
• B. 3
• C. 4
• D. 6

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We have a point \(P(1, 2)\) lying on the given circle. Any chord passing through \(P\) and another point \(Q(x_1, y_1)\) on the circle is bisected by the y-axis, meaning the midpoint of \(PQ\) lies on the line \(x = 0\).
Step 2: Key Formula or Approach:
If the midpoint \(M\) of \(PQ\) lies on the y-axis, its x-coordinate must be 0.
Using the midpoint formula: \(x_M = \frac{1 + x_1}{2} = 0\).
Step 3: Detailed Explanation:
Let's find the coordinates of \(Q\).
\[ \frac{1 + x_1}{2} = 0 \implies x_1 = -1 \] Since \(Q(-1, y_1)\) must also lie on the circle \(x^2 + y^2 + x - 3y = 0\), substitute \(x = -1\):
\[ (-1)^2 + y_1^2 + (-1) - 3y_1 = 0 \implies 1 + y_1^2 - 1 - 3y_1 = 0 \implies y_1(y_1 - 3) = 0 \] This gives two possible values for \(y_1\): \(y_1 = 0\) and \(y_1 = 3\).
Therefore, the other ends of the two chords are the points \(R(-1, 0)\) and \(S(-1, 3)\).
Now, we need to find the midpoint of the line segment \(RS\), which is \((\alpha, \beta)\):
\[ (\alpha, \beta) = \left( \frac{-1 + (-1)}{2}, \frac{0 + 3}{2} \right) = \left( -1, \frac{3}{2} \right) \] So, \(\alpha = -1\) and \(\beta = \frac{3}{2}\).
Step 4: Final Answer:
We must calculate the value of \(6(\alpha + \beta)\):
\[ 6(\alpha + \beta) = 6\left(-1 + \frac{3}{2}\right) = 6\left(\frac{1}{2}\right) = 3 \]