Question

If \( P \) is a point on the circle \( x^2 + y^2 = 4 \), \( Q \) is a point on the straight line \( 5x + y + 2 = 0 \) and \( x - y + 1 = 0 \) is the perpendicular bisector of \( PQ \), then 13 times the sum of abscissa of all such points \( P \) is _________.

Hint:

The image of a point \( (x, y) \) about the line \( y = x + c \) is simply \( (y-c, x+c) \). Similarly for \( y = -x + c \). Using this shortcut saves time during the exam.

Answer 1

Correct Answer: 20

Step 1: Understanding the Concept:
If \( L: x - y + 1 = 0 \) is the perpendicular bisector of \( PQ \), then point \( Q \) is the reflection (image) of point \( P \) with respect to line \( L \). Since \( Q \) lies on a given line, we can use the image formula to find a constraint on \( P \).

Step 2: Key Formula or Approach:
Image of \( P(x_1, y_1) \) about \( ax + by + c = 0 \) is \( Q(x_2, y_2) \) where:
\[ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2} \]

Step 3: Detailed Explanation:
Let \( P = (x_1, y_1) \). For line \( x - y + 1 = 0 \):
\[ \frac{x_2 - x_1}{1} = \frac{y_2 - y_1}{-1} = -2 \frac{x_1 - y_1 + 1}{1 + 1} = -(x_1 - y_1 + 1) \]
Solving for \( x_2 \) and \( y_2 \):
\( x_2 = x_1 - (x_1 - y_1 + 1) = y_1 - 1 \)
\( y_2 = y_1 + (x_1 - y_1 + 1) = x_1 + 1 \)
Since \( Q(y_1 - 1, x_1 + 1) \) lies on \( 5x + y + 2 = 0 \):
\[ 5(y_1 - 1) + (x_1 + 1) + 2 = 0 \implies x_1 + 5y_1 - 2 = 0 \implies x_1 + 5y_1 = 2 \]
Also, \( P \) lies on the circle \( x_1^2 + y_1^2 = 4 \).
We need the sum of abscissas (\( x_1 \)). Substitute \( y_1 = \frac{2 - x_1}{5} \) into the circle equation:
\[ x_1^2 + \left( \frac{2 - x_1}{5} \right)^2 = 4 \]
\[ 25x_1^2 + 4 - 4x_1 + x_1^2 = 100 \implies 26x_1^2 - 4x_1 - 96 = 0 \]
Dividing by 2: \( 13x_1^2 - 2x_1 - 48 = 0 \).
The sum of roots (sum of all possible abscissas of \( P \)) is \( -\frac{b}{a} = \frac{2}{13} \).
The required value is \( 13 \times (\text{sum of abscissas}) = 13 \times \frac{2}{13} = 2 \).

Step 4: Final Answer:
The final value is 2.