Question:
Let a circle pass through the origin and its center be the point of intersection of two mutually perpendicular lines \( x + (k-1)y + 3 = 0 \) and \( 2x + k2y - 4 = 0 \). If the line \( x - y + 2 = 0 \) intersects the circle at the points A and B, then \( (AB)^2 \) is equal to:
Let a circle pass through the origin and its center be the point of intersection of two mutually perpendicular lines \( x + (k-1)y + 3 = 0 \) and \( 2x + k2y - 4 = 0 \). If the line \( x - y + 2 = 0 \) intersects the circle at the points A and B, then \( (AB)^2 \) is equal to:
Updated On: Jun 6, 2026
- 10
- 27
- 18
- 34
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Equation of the circle.
The center of the circle lies at the intersection of the two mutually perpendicular lines. To find the center, we solve the system of equations: \[ x + (k-1)y + 3 = 0
2x + k2y - 4 = 0 \] By solving this system, we find the coordinates of the center \( (h, k) \).
Step 2: Use the equation of the line and circle.
Next, substitute the equation of the line \( x - y + 2 = 0 \) into the circle equation to find the points of intersection A and B, and calculate \( (AB)^2 \).
Final Answer: 18
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