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In the coordinate plane, point P has coordinates (a, b). Is point P inside the circle of radius 5 centered at the origin?
(1) a + b = 6
(2) |a| + |b| = 4
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Correct answer
B
Point P is inside the circle exactly when a² + b² < 25.
Statement (1): a + b = 6 is a line whose points range from inside to outside, (3,3) gives a² + b² = 18 (inside) while (6,0) gives 36 (outside). Straddles. Not sufficient.
Statement (2): |a| + |b| = 4 is a diamond (square rotated 45 degrees) with vertices at (±4, 0) and (0, ±4). The point on this diamond farthest from the origin is a vertex, where a² + b² = 16; the nearest points are the edge midpoints like (2, 2), where a² + b² = 8. So for every point on the diamond, 8 ≤ a² + b² ≤ 16, which is always below 25. P is therefore inside the circle no matter where it sits on the diamond, a determinate yes. Sufficient.
The 'line locates the point' trap (pick A or D) over-trusts the linear constraint, but a line crosses both inside and outside the circle. (2) gives an L1 ('taxicab') constraint, not the distance-squared directly, so a solver must find the diamond's farthest point from the origin (a² + b² ≤ 16) before concluding 'always inside', a solver who treats |a| + |b| = 4 as if it forced a single a² + b² value, or who can't bound it, may mis-key (E) or (C).
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