AMC 10 · 2023 · #10

Grade 10 geometry-2d
coordinate-geometrytangent-circlessystems-of-equationsslope-intercept convert-to-algebracoordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryslope-intercept
📏 Medium solution 💡 3 insights
Problem
Two circles sit in the plane. One has radius 4, its centre on the positive x-axis, touching the y-axis at the origin. The other has radius 10, its centre on the positive y-axis, touching the x-axis at the origin. They cross at two points. Find the slope of the line through those two points.

Pick an answer.

(A)
$\dfrac{2}{7}$
(B)
$\dfrac{3}{7}$
(C)
$\dfrac{2}{\sqrt{29}}$
(D)
$\dfrac{1}{\sqrt{29}}$
(E)
$\dfrac{2}{5}$
How to solve
Strategy Convert to Algebra

The problem is handed over in words about tangency and axes, so Tool #13 (Convert to Algebra) owns it: once each circle becomes an equation, everything after that is symbol pushing. Tool #1 (Draw a Diagram) does the first real work, because "tangent to the y-axis at the origin" is a picture statement, and reading the picture is what pins each center to a single point rather than a range. Tool #15 (Organize Information in More Ways) is the hinge — the standard form (x-h)²+(y-k)²=r² hides what is useful here, and expanding both circles into the form x²+y²=something linear lines the two equations up so that their curved parts are literally identical. Tool #16 (Change Focus) is the punchline: the question asks about a line through two points, and the reflex is to hunt down both points first. Chasing the line itself instead makes the two points unnecessary, and the whole problem collapses to one subtraction. Tool #7 (Identify Subproblems) then handles a small side job at the end: confirming that the two circles really do meet twice, so the line found by subtraction is a line through actual points.

1STEP 1

Pin down both centres

Tangency fixes each centre.

Circle 1 center = (4,0), r₁=4; Circle 2 center = (0,10), r₂=10
2STEP 2

Write each equation

Write each circle as an equation.

(x-4)²+y²=16 and x²+(y-10)²=100
3STEP 3

Expanding kills the constants

Expanding kills the constant terms.

x²+y²=8x and x²+y²=20y
4STEP 4

Subtract to get the line

Subtracting makes the squares vanish.

x²+y²=8x, x²+y²=20y ⟹ 8x=20y
5STEP 5

Find the crossings

Substitute to find both crossings.

x²+(2/5x)²=8x ⟹ 29/25x²=8x ⟹ x(29x-200)=0 ⟹ (x,y)=(0,0) or (200/29,80/29)
6STEP 6

Read the slope

The slope is two fifths.

8x=20y ⟹ y=8/20x=2/5x ⟹ m=2/5 (E)
Answer
2/5
Check the two points against the original circle equations, using no result from the solution. For (0,0): circle 1 gives (0-4)²+0²=16, matching r₁²=16, and circle 2 gives 0²+(0-10)²=100, matching r₂²=100. For (200/29,80/29): circle 1 gives (200/29-116/29)²+(80/29)²=(84²+80²)/841=(7056+6400)/841=13456/841=16, and circle 2 gives (200/29)²+(80/29-290/29)²=(40000+44100)/841=84100/841=100. Both points sit on both circles exactly, and the slope between them is 2/5. The circles genuinely have to cross twice, too: the distance between the centers (4,0) and (0,10) is √(16+100)=√(116)≈ 10.77, which lies strictly between 10-4=6 and 10+4=14, the condition for two intersection points. Two of the answer choices, 2/√(29)≈ 0.371 and 1/√(29)≈ 0.186, are traps built from that same √(116)=2√(29); the correct value 2/5=0.4 is close to the first one, so reducing 8/20 carefully rather than estimating matters here.
💡Key takeaway

Two circle equations both contain the same x²+y², so subtracting one from the other wipes out the curvature and leaves the straight line through their two crossing points.

  • Pin down both centers
  • Write each circle's equation
  • Expand; the constants cancel
  • Subtract to get the line
  • Find the two crossing points
  • Read the slope