Two circles with equal radii are intersecting at the points and . The tangent at the point to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these circles is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Let the two circles have equal radii r and centers C₁ and C₂. Both circles pass through (0,1) and (0,-1), so both centers lie on the perpendicular bisector of the segment joining these points, which is the x-axis. Let C₁ = (a, 0) and C₂ = (-a, 0) by symmetry.
The radius satisfies r² = a² + 1.
The tangent to the first circle at (0,1) is perpendicular to the radius from C₁ = (a, 0) to (0,1). The radius vector is (0-a, 1-0) = (-a, 1). The tangent at (0,1) has direction (1, a) (perpendicular to radius), so its equation passes through (0,1): parametrically (t, 1+at).
This tangent must pass through C₂ = (-a, 0). So: t = -a and 1 + a(-a) = 0 → 1 - a² = 0 → a = 1.
The distance between centers = 2a = 2.
The radius satisfies r² = a² + 1.
The tangent to the first circle at (0,1) is perpendicular to the radius from C₁ = (a, 0) to (0,1). The radius vector is (0-a, 1-0) = (-a, 1). The tangent at (0,1) has direction (1, a) (perpendicular to radius), so its equation passes through (0,1): parametrically (t, 1+at).
This tangent must pass through C₂ = (-a, 0). So: t = -a and 1 + a(-a) = 0 → 1 - a² = 0 → a = 1.
The distance between centers = 2a = 2.
[JEE Main 2019 · 11 Jan · Shift I]