Step 1: The given circle equation is \( x^2 + y^2 = 4 \). The equation of the line \( x + y = 1 \) will intersect the circle at two points, which we need to find.
Step 2: Solve the system of equations \( x + y = 1 \) and \( x^2 + y^2 = 4 \) to find the points of intersection A and B.
Step 3: The line perpendicular to \( AB \) passing through the midpoint of \( AB \) will intersect the circle again at points C and D. Use geometric properties of the circle and the perpendicular bisector to find the area of quadrilateral ABCD.
Step 4: After solving for the coordinates of points A, B, C, and D, calculate the area of quadrilateral ABCD, which evaluates to \( \sqrt{14} \). Thus, the correct answer is (1).
Let \( ABC \) be a triangle formed by the lines \( 7x - 6y + 3 = 0 \), \( x + 2y - 31 = 0 \), and \( 9x - 2y - 19 = 0 \).
Let the point \( (h, k) \) be the image of the centroid of \( \triangle ABC \) in the line \( 3x + 6y - 53 = 0 \). Then \( h^2 + k^2 + hk \) is equal to:
Let \( \overrightarrow{a} = i + 2j + k \) and \( \overrightarrow{b} = 2i + 7j + 3k \).
Let \[ L_1 : \overrightarrow{r} = (-i + 2j + k) + \lambda \overrightarrow{a}, \quad \lambda \in \mathbb{R} \] and \[ L_2 : \overrightarrow{r} = (j + k) + \mu \overrightarrow{b}, \quad \mu \in \mathbb{R} \] be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \overrightarrow{a} + \overrightarrow{b} \), then \( L_3 \) passes through the point: