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

Given the circle equation \(x^2 + y^2 = 4\), any point \(P(x_1, y_1)\) on the circle satisfies \(x_1^2 + y_1^2 = 4\). The line equation given is \(5x + y + 2 = 0\), and the perpendicular bisector of segment \(PQ\) is \(x - y + 1 = 0\).

Let's examine the properties of the perpendicular bisector: \(x - y + 1 = 0\) implies that the slope of the line is 1. For the line segment \(PQ\), if this is the perpendicular bisector, the slope of \(PQ\) would be \(-1\) due to the perpendicular relationship.

Solving the equation \(x - y + 1 = 0\):  \(x = y - 1\)

Substitute into the line equation for \(Q\): \(5(y - 1) + y + 2 = 0\) \(5y - 5 + y + 2 = 0\) \(6y - 3 = 0\) \(y = \frac{1}{2}\)

Thus, for point \(Q\), \(x = y - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\), giving \(Q\left(-\frac{1}{2}, \frac{1}{2}\right)\).

The midpoint of \(PQ\) lies on the bisector \(x - y + 1 = 0\): \(x - y + 1 = 0 \Rightarrow x = y - 1\)

Midpoint of \(PQ\): \(\left(\frac{x_1 + \left(-\frac{1}{2}\right)}{2}, \frac{y_1 + \frac{1}{2}}{2}\right)\) Substitute midpoint relation: \(\frac{x_1 - \frac{1}{2}}{2} = \frac{y_1 + \frac{1}{2}}{2} - 1\) or, \(x_1 - \frac{1}{2} = y_1 + \frac{1}{2} - 2\) \(x_1 = y_1 - \frac{3}{2}\)

For \(P(x_1, y_1)\) lying on the circle, substitute into the circle equation: \((y_1 - \frac{3}{2})^2 + y_1^2 = 4\) This simplifies to: \(y_1^2 - 3y_1 + \frac{9}{4} + y_1^2 = 4\) \(2y_1^2 - 3y_1 + \frac{9}{4} = 4\) \(2y_1^2 - 3y_1 - \frac{7}{4} = 0\)

Multiply through by 4 to clear fractions: \(8y_1^2 - 12y_1 - 7 = 0\) Using the quadratic formula: \(y_1 = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 8 \cdot (-7)}}{2 \cdot 8}\) \(y_1 = \frac{12 \pm \sqrt{144 + 224}}{16}\) \(y_1 = \frac{12 \pm \sqrt{368}}{16}\) \(y_1 = \frac{12 \pm 2\sqrt{92}}{16}\) \(y_1 = \frac{6 \pm \sqrt{92}}{8}\) \(\sqrt{92} = 2\sqrt{23}\), yields: \(y_1 = \frac{6 \pm 2\sqrt{23}}{8}\) \(y_1 = \frac{3 \pm \sqrt{23}}{4}\)

For each \(y_1\), calculate \(x_1 = y_1 - \frac{3}{2}\):

1. Calculate for \(y_1 = \frac{3 + \sqrt{23}}{4}\): \(x_1 = \frac{3 + \sqrt{23}}{4} - \frac{3}{2}\) \(x_1 = \frac{3 + \sqrt{23} - 6}{4}\) \(x_1 = \frac{\sqrt{23} - 3}{4}\)

2. Calculate for \(y_1 = \frac{3 - \sqrt{23}}{4}\): \(x_1 = \frac{3 - \sqrt{23}}{4} - \frac{3}{2}\) \(x_1 = \frac{3 - \sqrt{23} - 6}{4}\) \(x_1 = \frac{-\sqrt{23} - 3}{4}\)

The sum of abscissas: \(\frac{\sqrt{23} - 3}{4} + \frac{-\sqrt{23} - 3}{4} = \frac{-6}{4} = -\frac{3}{2}\)

Thus, multiplying with 13: \(13 \cdot \left(-\frac{3}{2}\right) = -\frac{39}{2}\)