AMC 10 · 2017 · #6

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
circle-equationmidpoint-formulapythagorean-theorem coordinate-geometryidentify-subproblems ↑ Prerequisites: circle-equationmidpoint-formula
📏 Medium solution 💡 2 insights
Problem
A circle is drawn so that the segment joining the origin and the point (8,6) is a diameter. Because the origin sits on the x-axis, that point is already one crossing of the x-axis, and the circle crosses the x-axis once more. Find the x-coordinate of that other crossing.

Pick an answer.

(A)
$4\sqrt{2}$
(B)
6
(C)
$5\sqrt{2}$
(D)
8
(E)
$6\sqrt{2}$
How to solve
Strategy Introduce a Variable

The unknown point is pinned to the x-axis, so calling it (x,0) turns two unknown coordinates into one. That single name is what makes the rest mechanical: "the point is on the circle" becomes "the point is exactly one radius from the center," which is a distance statement I can write as an equation in x. Before that equation can be written I need the center and the radius, so I split the work into three small jobs: find the center, find the radius, then solve for x. Sketching the circle with its center marked keeps the picture honest and shows why the x-axis meets it twice rather than once. At the end the two candidate values of x must be sorted out, and one of them is a possibility to eliminate rather than the answer.

1STEP 1

Find the center from the diameter

The diameter's midpoint is the centre.

((0+8)/2, (0+6)/2) = (4,3)
2STEP 2

Get the radius

Measuring centre to endpoint gives the radius.

r = √(4²+3²) = √(16+9) = √(25) = 5
3STEP 3

Name the unknown point

A point on the axis has y equal to zero.

(x-4)² + 3² = 5²
4STEP 4

Solve for the horizontal leg

Undoing the square gives two roots.

(x-4)² = 16 → x-4 = 4 or x-4 = -4 → x = 8 or x = 0
5STEP 5

Rule out the given crossing

Discarding the known one leaves 8.

(8-4)² + (0-3)² = 16 + 9 = 25 = 5²
Answer
8
The center is at x = 4 with radius 5, so every point of the circle has an x-coordinate between -1 and 9; the value 8 sits inside that band, and the two crossings 0 and 8 are symmetric about x = 4, exactly as a chord of a circle must be. A second route that never computes the center or the radius confirms it. A point P on a circle whose diameter is AB always sees that diameter at a right angle, so with A = (0,0), B = (8,6), and P = (x,0), the segment PA runs along the x-axis and is therefore horizontal. For the angle at P to be a right angle, PB must be vertical, which forces P to sit directly below B, giving x = 8. Two independent routes landing on 8 makes the answer solid.
💡Key takeaway

A diameter hands you the center for free — average its endpoints, measure out one radius with the Pythagorean theorem, and remember that a squared quantity has two roots because the line really does cut the circle twice.

  • Find the center from the diameter
  • Get the radius
  • Name the unknown point
  • Solve for the horizontal leg
  • Rule out the given crossing

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