Coordinate Geometry — Hard JEE math MCQ
**[JEE Mains 2026]** If two circles x² + y² - 4x + 2y - 4 = 0 and (x-1)² + (y-4)² = r² intersect at two distinct points, what is the range of r?
- A. 2 < r < 8
- B. 1 < r < 7
- C. 0 < r < 6
- D. 3 < r < 9
Solution
First circle: center C₁(2, -1), radius r₁ = √(4+1+4) = 3.
Second circle: center C₂(1, 4), radius r₂ = r.
Distance between centers: d = √((2-1)² + (-1-4)²) = √(1+25) = √26 ≈ 5.1.
For intersection: |r₁ - r₂| < d < r₁ + r₂.
|3 - r| < √26 < 3 + r.
From √26 < 3 + r: r > √26 - 3 ≈ 2.1.
From |3 - r| < √26: -√26 < 3 - r < √26, so 3 - √26 < r < 3 + √26.
Range: approximately 2 < r < 8.