JEE Mains · Maths · STD 11 - 9. straight line
Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines \(x + (k-1)y + 3 = 0\) and \(2x + k^2 y - 4 = 0\). If the line \(x - y + 2 = 0\) intersects the circle at the points A and B, then \((AB)^2\) is equal to:
- A \(10\)
- B \(27\)
- C \(18\)
- D \(34\)
Answer & Solution
Correct Answer
(C) \(18\)
Step-by-step Solution
Detailed explanation
The given lines are \(x + (k-1)y + 3 = 0\) and \(2x + k^2 y - 4 = 0\). Since they are mutually perpendicular, the product of their slopes is \(-1\): \(\left(\dfrac{-1}{k-1}\right) \left(\dfrac{-2}{k^2}\right) = -1\) \(\dfrac{2}{k^2(k-1)} = -1 \Rightarrow k^3 - k^2 + 2 = 0\) By…
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