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On an engineering grid, a circular sensor zone is drawn so that the points A(1, 2) and B(9, 8) are the two endpoints of a diameter of the zone's boundary circle. A repair access point sits on the boundary circle at height y = 1, on the side of the circle to the right of the center. What is the x-coordinate of the repair access point?
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Correct answer
C
The two given points are the ends of a diameter, so the center of the circle is their midpoint and the radius is half the distance between them.
Center (midpoint of A(1, 2) and B(9, 8)): ((1 + 9)/2, (2 + 8)/2) = (5, 5). Radius (half the diameter): the distance from A to B is √((9 − 1)² + (8 − 2)²) = √(64 + 36) = √100 = 10, so the radius is 10 ÷ 2 = 5. The boundary circle is (x − 5)² + (y − 5)² = 25.
Now set y = 1 to find the access point on the circle at that height: (x − 5)² + (1 − 5)² = 25, so (x − 5)² + 16 = 25, giving (x − 5)² = 9 and x − 5 = ±3. The two solutions are x = 8 and x = 2. The access point sits to the right of the center, so x = 8. The answer is (C).
As a quick geometric check: the point at y = 1 is 4 units below the center, and the radius is 5, so the horizontal reach from the center to the boundary at that height is √(5² − 4²) = √9 = 3. To the right of the center that lands at 5 + 3 = 8, confirming (C).
Why each wrong choice tempts you. (A) 2 is the other root of (x − 5)² = 9; it is the point to the left of the center, so it is ruled out by the stated right-of-center condition. (B) 3 comes from solving x − 5 = 3 and stopping there, forgetting to add the center's x-coordinate 5 to get x = 8. (D) 9 lifts the x-coordinate of endpoint B straight from the stem, treating a given diameter end as if it were the requested point rather than a new point at y = 1. (E) 10 drops the y = 1 height altogether, solving (x − 5)² = 25 for the rightmost point of the whole circle instead of the point at the required height. The two slips most likely to feel correct are halving the diameter to get the radius and then reading the right branch of the two roots, which is exactly where (A) and (B) catch a rushed solver.
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