The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle is
- A.(3/2, 1/2)
- B.(1/2, 3/2)
- C.(1/2, 1/2)
- D.
Show answer and solution
Correct option: D
Solution
The center must lie on the perpendicular bisector of the chord joining (0,0) and (1,0), which is the vertical line x = 1/2.
So center is (1/2, k) for some k. The radius r = distance from center to (0,0) = √(1/4 + k²).
Since the circle touches x² + y² = 9 (radius 3, center at origin), the distance between centers equals 3 ± r (internal or external tangency):
√(1/4 + k²) = |3 - √(1/4 + k²)| or = 3 + √(1/4 + k²).
For internal tangency: distance between centers = 3 - r, so √(1/4 + k²) = 3 - √(1/4 + k²), giving 2√(1/4 + k²) = 3, so √(1/4 + k²) = 3/2, thus 1/4 + k² = 9/4, k² = 2, k = ±√2.
The center is (1/2, -√2), matching option 4.
So center is (1/2, k) for some k. The radius r = distance from center to (0,0) = √(1/4 + k²).
Since the circle touches x² + y² = 9 (radius 3, center at origin), the distance between centers equals 3 ± r (internal or external tangency):
√(1/4 + k²) = |3 - √(1/4 + k²)| or = 3 + √(1/4 + k²).
For internal tangency: distance between centers = 3 - r, so √(1/4 + k²) = 3 - √(1/4 + k²), giving 2√(1/4 + k²) = 3, so √(1/4 + k²) = 3/2, thus 1/4 + k² = 9/4, k² = 2, k = ±√2.
The center is (1/2, -√2), matching option 4.
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