AMC 10 · 2019 · #20

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

pythagorean-theoremcoordinate-geometrychord-perpendicular-from-centerperpendicular-bisectorarea-circles identify-subproblems ↑ Prerequisites: pythagorean-theoremcoordinate-geometryarea-circles

📏 Long solution 💡 4 insights

Problem

The points (6,13) and (12,11) lie on a circle. The tangent lines at those two points meet at a single point on the x-axis. Find the area of the circle.

Pick an answer.

(A)

$\frac{83\pi}{8}$

(B)

$\frac{21\pi}{2}$

(C)

$\frac{85\pi}{8}$

(D)

$\frac{43\pi}{4}$

(E)

$\frac{87\pi}{8}$

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): the picture (circle, two tangent points, external point on the x-axis) clarifies that |PA| = |PB| and that △ OAP is right-angled at A. Tool #7 (Subproblems): split into (a) find P on the x-axis from |PA| = |PB|, (b) compute |PA|, (c) use right-triangle/Ptolemy to relate r, |PA|, |PO|, (d) solve for r². Tool #13 (Algebra): each subproblem reduces to a short algebraic step (linear equation for t, distance squared, Pythagorean relation, Ptolemy).

1STEP 1

Mark the right angles

A tangent meets its radius at a right angle.

P = (t, 0), OA ⊥ PA, OB ⊥ PB

2STEP 2

Locate the meeting point

Equal tangent lengths pin that point.

(t-6)² + 169 = (t-12)² + 121 → 12t = 60 → t = 5

3STEP 3

Find the tangent length

Compute the tangent length from there.

|PA|² = |PB|² = 170

4STEP 4

Measure the chord and midpoint

Measure the chord joining the touch points.

|AB| = 2√(10), |PM| = 4√(10)

5STEP 5

Use area to relate them

Writing one area two ways gives an equation.

2 √(170) · r = PO · 2√(10) → PO = r√(17)

6STEP 6

Solve for the radius squared

Solving gives the radius squared.

17 r² = r² + 170 → r² = 85/8

7STEP 7

Read the area

The area is eighty-five pi over eight.

Area = π r² = (85 π)/8

Answer

85π/8

Verify by independent coordinate computation. The center O lies on the perpendicular bisector of AB, which has slope -1/3, so the bisector has slope 3 and passes through M = (9, 12): y = 3x - 15. The center also lies on line PM from P = (5, 0) through M = (9, 12): that line is y = 3(x - 5) = 3x - 15 — same line ✓ (as it should be, since the locus of points equidistant from the two tangents through A, B is exactly the line through P and M). So O is on y = 3x - 15, and additionally on the line through A perpendicular to PA. PA direction is (1, 13), so the line through A = (6, 13) perpendicular to PA has direction (-13, 1): parametrically (6 - 13s, 13 + s). Plug into y = 3x - 15: 13 + s = 3(6 - 13s) - 15 = 3 - 39s, so 40s = -10, s = -1/4. Then O = (6 + 13/4, 13 - 1/4) = (37/4, 51/4). Now r² = (37/4 - 6)² + (51/4 - 13)² = (13/4)² + (-1/4)² = (169 + 1)/16 = 170/16 = 85/8 ✓. Same value. Area = 85π/8.

💡Key takeaway

This AMC 12 problem only needs Grade 8 Pythagoras and the circle's basic 'radius ⟂ tangent' fact — solve the linear equation |PA| = |PB| to find P = (5, 0), get |PA|² = 170, then 16 r² = 170 from a right-triangle/Ptolemy relation, so the area is (85 π)/8.