Question

A circle $x^2 + y^2 + x - 3y = 0$ passes through $P(1, 2)$. If 2 chords (PS \& PR) drawn from P are bisected by $y$-axis, then mid point of RS is $(\alpha, \beta)$, find $6(\alpha + \beta)$

• A. 3
• B. 5
• C. 4
• D. 7

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A chord $PS$ from point $P(x_1, y_1)$ to $S(x_2, y_2)$ has its midpoint at $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
If the midpoint lies on the $y$-axis, its $x$-coordinate must be zero.
Step 2: Key Formula or Approach:
1. Find the $x$-coordinates of points $R$ and $S$ using the midpoint condition.
2. Since $R$ and $S$ lie on the circle, find their $y$-coordinates by substituting into the circle equation.
3. Calculate the midpoint of $RS$ to find $(\alpha, \beta)$.
Step 3: Detailed Explanation:
1. Let $S = (x_S, y_S)$ and $R = (x_R, y_R)$.
Midpoint of $PS$ on $y$-axis $\implies \frac{1 + x_S}{2} = 0 \implies x_S = -1$.
Midpoint of $PR$ on $y$-axis $\implies \frac{1 + x_R}{2} = 0 \implies x_R = -1$.
2. Substitute $x = -1$ into circle equation $x^2 + y^2 + x - 3y = 0$:
$(-1)^2 + y^2 + (-1) - 3y = 0 \implies y^2 - 3y = 0 \implies y(y - 3) = 0$.
Thus $y = 0$ or $y = 3$.
So the points are $S(-1, 0)$ and $R(-1, 3)$.
3. Find midpoint $(\alpha, \beta)$ of $RS$:
$\alpha = \frac{-1 + (-1)}{2} = -1$.
$\beta = \frac{0 + 3}{2} = 1.5$.
4. Calculate $6(\alpha + \beta)$:
$6(-1 + 1.5) = 6(0.5) = 3$.
Step 4: Final Answer:
The value of $6(\alpha + \beta)$ is 3.