AMC 10 · 2017 · #9

Grade 10 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
circle-equationsystems-of-equationscoordinate-geometry convert-to-algebraidentify-subproblems ↑ Prerequisites: circle-equationsystems-of-equations
📏 Medium solution 💡 2 insights
Problem
Two circles are given by centre and radius: one centred at (-10, -4) with radius 13, the other centred at (3, 9) with radius root 65. They cross at two points, and the line through those points can be written as x plus y equals c. Find the number c.

Pick an answer.

(A)
3
(B)
$3\sqrt{3}$
(C)
$4\sqrt{2}$
(D)
6
(E)
$\frac{13}{2}$
How to solve
Strategy Convert to Algebra

Hunting for the two intersection points one at a time means solving a system with squares in it, which is messy. Instead, turn each circle into an equation (Convert to Algebra) and notice that both equations hold at both crossing points. Subtracting one equation from the other reorganizes the same information (Organize Information in More Ways) into a single equation with no squared terms left, and that leftover equation is exactly the line asked about. A quick sketch (Draw a Diagram) first confirms that the two circles genuinely cross, and splitting the work into 'do they cross', 'write the equations', 'subtract', 'solve' (Identify Subproblems) keeps each step small.

1STEP 1

Check the circles really cross

The centre distance confirms they really do cross.

d = √(13² + 13²) = 13√(2) ≈ 18.38, 4.94 ≈ 13 - √(65) ＜ 13√(2) ＜ 13 + √(65) ≈ 21.06
2STEP 2

Write each circle as an equation

Each circle becomes an equation.

(x+10)² + (y+4)² = 169 (x-3)² + (y-9)² = 65
3STEP 3

Both equations hold at the crossings

At a crossing both equations hold.

(x+10)²+(y+4)²-169 = 0 (x-3)²+(y-9)²-65 = 0
4STEP 4

Subtract to cancel the squares

Subtracting makes every squared term vanish.

(x²+y²+20x+8y+116) - (x²+y²-6x-18y+90) = 169 - 65 → 26x + 26y + 26 = 104
5STEP 5

Read off c

The leftover linear equation reads off 3.

26x + 26y + 26 = 104 → 26(x+y) = 78 → x + y = 3 → c = 3
Answer
3
Produce the actual crossing points and test them. The point (2, 1) sits 12 right and 5 up from (-10, -4), and 5-12-13 is a right triangle, so its distance to the first center is exactly 13; it also sits 1 left and 8 down from (3, 9), and √(1² + 8²) = √(65), so it is on the second circle too. The point (-5, 8) passes both tests as well: 5 right and 12 up from (-10, -4) gives 13, and 8 left and 1 down from (3, 9) gives √(65). Two circles cannot meet more than twice, so these are the two intersection points, and both give the same sum, 2 + 1 = 3 and (-5) + 8 = 3. The direction also checks out: the segment joining the centers has slope 1, a common chord is perpendicular to that segment, and every line of the form x + y = c has slope -1. So c = 3, which is choice (A).
💡Key takeaway

Both circle equations are true at the crossing points, so subtracting one from the other kills the squared terms and leaves exactly the line through those points.

  • Check the circles really cross
  • Write each circle as an equation
  • Both equations hold at the crossings
  • Subtract to cancel the squares
  • Read off c

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