Question

The line \( x + y + 1 = 0 \) intersects the circle \( x^2 + y^2 - 4x + 2y - 4 = 0 \) at the points A and B. If \( M(a, b) \) is the midpoint of AB, then \( a - b = \)?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When solving for intersections of a line and a circle, use substitution to eliminate one variable, and solve the resulting quadratic equation for the points of intersection. The midpoint formula can then be applied to find the center.

Answer 1

The Correct Option is D

We are given the line \( x + y + 1 = 0 \), which can be rewritten as \( y = -x - 1 \). We will substitute this expression for \( y \) into the equation of the circle \( x^2 + y^2 - 4x + 2y - 4 = 0 \). Substituting \( y = -x - 1 \) into the circle's equation: \[ x^2 + (-x - 1)^2 - 4x + 2(-x - 1) - 4 = 0 \] Simplifying the terms: \[ x^2 + (x^2 + 2x + 1) - 4x - 2x - 2 - 4 = 0 \] \[ 2x^2 - 4x - 5 = 0 \] Dividing the entire equation by 2: \[ x^2 - 2x - \frac{5}{2} = 0 \] We solve this quadratic equation using the quadratic formula: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-\frac{5}{2})}}{2(1)} \] \[ x = \frac{2 \pm \sqrt{4 + 10}}{2} = \frac{2 \pm \sqrt{14}}{2} \] So the two values of \(x\) are: \[ x_1 = \frac{2 + \sqrt{14}}{2}, \quad x_2 = \frac{2 - \sqrt{14}}{2} \] Now, to find the corresponding \(y\)-coordinates, substitute these \(x\)-values back into the equation \(y = -x - 1\). For the first \(x\)-value: \[ y_1 = -\left(\frac{2 + \sqrt{14}}{2}\right) - 1 = \frac{-2 - \sqrt{14} - 2}{2} = \frac{-4 - \sqrt{14}}{2} \] Similarly for the second \(x\)-value: \[ y_2 = -\left(\frac{2 - \sqrt{14}}{2}\right) - 1 = \frac{-2 + \sqrt{14} - 2}{2} = \frac{-4 + \sqrt{14}}{2} \] Now, the coordinates of the midpoint \(M(a,b)\) of AB are given by: \[ a = \frac{x_1 + x_2}{2}, \quad b = \frac{y_1 + y_2}{2} \] We can directly calculate \(a - b\) to find that the result is: \[ a - b = \frac{1}{2} \] Thus, the answer is \( \boxed{3} \).