AMC 10 · 2009 · #16

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

tangent-circlescoordinate-geometryquadratic-equations convert-to-algebracompleting-the-square ↑ Prerequisites: coordinate-geometry

📏 Long solution 💡 3 insights

Problem

A circle touches both positive axes and is externally tangent to a fixed circle. More than one such circle exists. Find the sum of all possible radii.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable): a circle in the plane normally needs three numbers, but the double tangency to the axes collapses all three into one — the radius r — because the center is forced to (r,r). Tool #13 (Convert to Algebra): external tangency is a sentence about touching, and the distance formula turns it into a single equation in r. Tool #15 (Organize Information in More Ways): completing the square rearranges that equation so the two roots appear as a matched pair around one center value, which is exactly the structure the final question asks about. Tool #3 (Eliminate Possibilities): the equation may produce roots that no circle can realize, so each root has to be tested against the geometry before it is allowed to count. Tool #16 (Change Focus / Count the Complement): the question wants the sum, not the radii, and a sum can be read off a quadratic's coefficients once the roots are known to be legitimate.

1STEP 1

Name the radius, pin the center

Touching both axes puts the centre on the diagonal.

|x₀|=|y₀|=r, x₀ ＞ 0, y₀ ＞ 0 ⟹ C=(r,r)

2STEP 2

Turn tangency into one equation

External tangency becomes one equation.

CA = r + 1

3STEP 3

Compute the distance and expand

Expanding gives a quadratic in the radius.

(r-3)²+r²=(r+1)² ⟺ 2r²-6r+9=r²+2r+1 ⟺ r²-8r+8=0

4STEP 4

Complete the square

Completing the square gives two roots.

r²-8r+16=8 ⟹ (r-4)²=8 ⟹ r=4± 2√(2)

5STEP 5

Check both roots are real circles

Both are positive, so both are real circles.

4-2√(2)≈ 1.17 ＞ 0, 4+2√(2)≈ 6.83 ＞ 0

6STEP 6

Add the radii

Adding them gives 8, choice (D).

(4-2√(2))+(4+2√(2))=8 → (D)

Answer

Verify both circles numerically. For r=4-2√(2)≈ 1.1716 the center is (1.1716,1.1716) and CA=√((1.1716-3)²+1.1716²)=√(4.7157)≈ 2.1716, which equals r+1. For r=4+2√(2)≈ 6.8284, CA=√(3.8284²+6.8284²)=√(61.284)≈ 7.8284, again r+1. Both configurations exist. The sizes match the picture too: the fixed circle sits on the x-axis spanning 2 ≤ x ≤ 4, so one solution circle is small and nestles in the corner between the origin and it, while the other is large and sweeps around past it. It is also worth testing whether the word externally is doing hidden work: internal tangency would require CA=|r-1|, giving r²-4r+8=0 with discriminant 16-32=-16 ＜ 0, so no internally tangent circle exists to be wrongly included or excluded — the configuration really is forced. Finally 8 exceeds either radius alone, as any sum of two positive radii must, and the distractor 4 is exactly the midpoint of the two roots, which is what a solver reports after finding only one root or stopping at the center of symmetry in (r-4)²=8.

💡Key takeaway

A circle tucked into the corner of both axes has its center at (r,r), so one number describes it completely — but before you add up the roots of the quadratic, make sure every root can actually be drawn.

Name the radius, pin the center
Turn tangency into one equation
Compute the distance and expand
Complete the square
Check both roots are real circles
Add the radii