JEE Main 2020 · 2 September · morning · Numerical answer · +4 / −0
The number of integral values of k for which the line, 3x + 4y = k intersects the circle,
x² + y² – 2x – 4y + 4 = 0 at two distinct points is ______.
Answer and solution
Answer 9
Solution
Maths sir · 4 steps
A line cuts a circle at two points exactly when its distance from the centre is less than the radius.
- Centre and radius. Complete the squares: (x-1)² + (y-2)² = 1 + 4 - 4 = 1. So the centre is (1, 2) and the radius is 1.
- Distance condition. For two distinct points, the perpendicular distance from (1, 2) to 3x + 4y - k = 0 must be strictly less than 1.
(|3(1) + 4(2) - k|)/√(3² + 4²) < 1 ⇒ |11 - k| < 5
- Solve for k. |11 - k| < 5 means -5 < 11 - k < 5, that is 6 < k < 16. The integers in this range are 7, 8, …, 15.
- So. There are 15 - 7 + 1 = 9 integral values of k.
The common slip: 11. Using ≤ instead of < also counts k = 6 and k = 16, where the line only touches the circle, and gives 11.
The idea to keep. Line versus circle: compare the distance from the centre with the radius. Less means two points, equal means tangent, more means no contact. Keep the inequality strict for distinct points.
Answer from the NTA final answer key. Solution by PrepWiser.