AMC 10 · 2002 · #18
Grade 10 geometry-2dPick an answer.
The two equations hide a picture, so Tool #1 (Draw a Diagram) comes first: read off both centres and radii and see that the circles sit apart from each other on the x-axis. Tool #17 (Visualize Spatial Relationships) then answers the question the algebra will not — a common tangent line can leave both centres on the same side of it, or separate them, and those are genuinely different segments. Tool #7 (Identify Subproblems) handles each configuration by folding the two perpendicular radii into a single right triangle whose hypotenuse is the fixed centre distance. Finally the word "shortest" is Tool #14 (Extreme Principle) — both configurations must be computed and compared, not just one of them.
Read centres and radii off the equations
Reading the equations gives radii 6 and 9 with centres 25 apart, and 25 exceeds 6+9 so the circles are separated.
A circle's equation carries its centre and radius in plain sight, so the picture is one reading away.
10.G-GPE.A.1Draw A DiagramTangency creates two right angles
A radius meets its tangent at a right angle, so the two radii are parallel — pointing the same way or opposite ways.
Two segments perpendicular to the same line must be parallel, so the only real choice left is which side of the line each centre takes.
Each radius drawn to a point of tangency stands square on the tangent line, so the two radii are parallel.
▸ Why?
A line that only grazes a circle at one point meets the radius drawn there at a right angle.
▸ Why?
Two lines standing square on the same line can never meet, so they keep a constant gap all the way along.
Fold each case into a right triangle
Laying the tangent on an axis makes each case a right triangle with hypotenuse 25 and vertical offset 9-6 or 9+6.
Putting the tangent line on the axis turns both perpendicular radii into vertical offsets, and the fixed centre distance becomes one hypotenuse.
8.G.B.7Identify SubproblemsSolve both and take the shorter
The crossing case subtracts the larger square, leaving L² = 625-225 = 400, so the shortest is 20, choice (C).
The hypotenuse is locked at 25, so the bigger the perpendicular leg, the less length is left for the segment along the line.
8.EE.A.2Extreme PrincipleA tangent meets the radius at a right angle, so slide both radii onto one right triangle with the centre distance as hypotenuse — and remember to check both ways the line can pass.
- Read centres and radii off the equations
- Tangency creates two right angles
- Fold each case into a right triangle
- Solve both and take the shorter