The locus of the centre of circle which touches externally and also touches X-axis, is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
Let the moving circle have centre (h, k) and radius r.
Since it touches the X-axis, r = |k|. For it to rest on or above the x-axis in the usual sense, we consider two cases.
Case 1: k ≥ 0 (centre above or on x-axis, so r = k)
External tangency with x² + (y-1)² = 1 (centre (0,1), radius 1):
Distance between centres = sum of radii
√(h² + (k-1)²) = k + 1
h² + (k-1)² = (k+1)²
h² = (k+1)² - (k-1)² = 4k
So x² = 4y, valid for y = k ≥ 0.
Case 2: k < 0 (centre below x-axis, r = -k)
External tangency condition:
√(h² + (k-1)²) = -k + 1 = 1 - k
h² + (k-1)² = (1-k)²
h² = 0 → h = 0
So the locus is the single line x = 0 (i.e., the y-axis) for all k < 0, giving the set {(0, y) : y < 0}.
Combining both cases, the locus is {x² = 4y, y ≥ 0} ∪ {(0, y), y < 0}, which is option 1.
Since it touches the X-axis, r = |k|. For it to rest on or above the x-axis in the usual sense, we consider two cases.
Case 1: k ≥ 0 (centre above or on x-axis, so r = k)
External tangency with x² + (y-1)² = 1 (centre (0,1), radius 1):
Distance between centres = sum of radii
√(h² + (k-1)²) = k + 1
h² + (k-1)² = (k+1)²
h² = (k+1)² - (k-1)² = 4k
So x² = 4y, valid for y = k ≥ 0.
Case 2: k < 0 (centre below x-axis, r = -k)
External tangency condition:
√(h² + (k-1)²) = -k + 1 = 1 - k
h² + (k-1)² = (1-k)²
h² = 0 → h = 0
So the locus is the single line x = 0 (i.e., the y-axis) for all k < 0, giving the set {(0, y) : y < 0}.
Combining both cases, the locus is {x² = 4y, y ≥ 0} ∪ {(0, y), y < 0}, which is option 1.
[AIEEE 2005]