If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then the length (in cm) of their common chord is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
Let the two circles have radii r₁ = 5 and r₂ = 12, intersecting at 90°. At the intersection point, the tangents to the two circles are perpendicular, which means the radii drawn to that point are also perpendicular to each other.
So the triangle formed by the two centers and the intersection point is right-angled, giving the distance between centers: d = √(5² + 12²) = √169 = 13.
For the common chord, use the formula for the length of the common chord. The common chord is the radical axis segment. Using the standard result: if the half-length of the common chord is h, then for each circle, h² = r₁² - a² where a is the distance from center 1 to the chord.
From center 1: a = (d² + r₁² - r₂²)/(2d) = (169 + 25 - 144)/26 = 50/26 = 25/13.
Half-chord length = √(r₁² - a²) = √(25 - 625/169) = √((4225 - 625)/169) = √(3600/169) = 60/13.
Full common chord = 2 × (60/13) = 120/13.
So the triangle formed by the two centers and the intersection point is right-angled, giving the distance between centers: d = √(5² + 12²) = √169 = 13.
For the common chord, use the formula for the length of the common chord. The common chord is the radical axis segment. Using the standard result: if the half-length of the common chord is h, then for each circle, h² = r₁² - a² where a is the distance from center 1 to the chord.
From center 1: a = (d² + r₁² - r₂²)/(2d) = (169 + 25 - 144)/26 = 50/26 = 25/13.
Half-chord length = √(r₁² - a²) = √(25 - 625/169) = √((4225 - 625)/169) = √(3600/169) = 60/13.
Full common chord = 2 × (60/13) = 120/13.
[JEE Main 2019 · 12 April · Shift I]