Prove that any two circles are similar using a translation followed by a dilation

Learning goal: Prove that all circles are similar using transformational arguments: show that for any two circles there is a translation followed by a dilation that maps the first circle exactly onto the second, and explain why that sequence of transformations is what "similar" means.

What similarity means in this course. Two figures are similar when there is a sequence of rigid motions (translations, rotations, reflections) and dilations that maps one figure onto the other. A rigid motion preserves every distance. A dilation with center O and scale factor k > 0 fixes the center O and sends each other point P to the point P′ on ray OP with OP′ = k·OP; it multiplies every distance by k and keeps every angle the same. So proving two figures similar is not a matter of measuring; it is a matter of producing the transformations and checking that the image of the first figure is the second figure, no more and no less.

What a circle is. A circle with center C and radius r is the set of all points whose distance from C is exactly r. Two things follow. A point is on the circle if and only if its distance from the center is r. And two circles with the same center and the same radius are the same circle, because they are the same set of points. Those "if and only if" statements are what let a transformation argument close: to show that the image of circle 1 is circle 2, we show that a point is on the image exactly when its distance from the second center is the second radius.

The claim. Let circle 1 have center A and radius r₁, and let circle 2 have center B and radius r₂. Then there is a translation followed by a dilation that maps circle 1 onto circle 2. Because a translation is a rigid motion and a dilation is a similarity transformation, circle 1 and circle 2 are similar. Since A, B, r₁ and r₂ were arbitrary, every pair of circles is similar; in short, all circles are similar.

Step one: translate. Use the translation T that sends A to B. A translation slides every point by the same vector, so it preserves distance: if P is any point, the distance from T(P) to T(A) = B equals the distance from P to A. Therefore the image of circle 1 under T is the set of points at distance r₁ from B, which is the circle with center B and radius r₁. Both directions of the "if and only if" hold: any point at distance r₁ from B is the image of a point at distance r₁ from A, because the translation can be undone.

Step two: dilate. Now apply the dilation D with center B and scale factor k = r₂/r₁. (Because radii are positive, k is a positive number, and if the circles happen to have the same radius then k = 1 and the dilation changes nothing.) A dilation centered at B multiplies the distance from B to any point by k. So a point at distance r₁ from B is sent to a point on the same ray at distance k·r₁ = (r₂/r₁)·r₁ = r₂ from B. The image of the circle with center B and radius r₁ is therefore the set of points at distance r₂ from B, which is circle 2. Again the reverse direction holds: every point at distance r₂ from B is the image of the point on the same ray at distance r₁, because a dilation with factor k can be undone by a dilation with factor 1/k.

Why this proves similarity. The composition "translate A to B, then dilate about B by r₂/r₁" maps circle 1 onto circle 2 exactly. That composition is a rigid motion followed by a dilation, which is exactly the definition of a similarity transformation. So circle 1 is similar to circle 2. Nothing in the argument depended on which circles were chosen, so the argument proves that all circles are similar.

A worked model with coordinates. Circle 1: center A(2, 3), radius 4. Circle 2: center B(−1, 5), radius 6. The translation that sends A to B moves every point 3 units left and 2 units up: (x, y) → (x − 3, y + 2). The scale factor is k = 6/4 = 3/2, and the dilation is centered at B(−1, 5). Follow one point of circle 1, say P(6, 3), which is on circle 1 because its distance from A is 4. The translation sends P to (3, 5). The dilation about B by 3/2 sends (3, 5) to (−1 + (3/2)(3 − (−1)), 5 + (3/2)(5 − 5)) = (−1 + 6, 5) = (5, 5). Check: the distance from (5, 5) to B(−1, 5) is 6, so the image point lies on circle 2, exactly as the proof says. Follow a second point, Q(2, 7), the top of circle 1: translation gives (−1, 9), and the dilation gives (−1 + (3/2)(0), 5 + (3/2)(9 − 5)) = (−1, 11), which is at distance 6 from B. Every point of circle 1 lands on circle 2 in the same way.

The misconception: "similar" needs matching shapes with different sizes, and circles are all the same shape anyway. Students sometimes say the statement is obvious and does not need proof, or the opposite, that circles cannot be "similar" because similarity is about triangles. Both miss the point. In this course "similar" has a precise meaning about transformations, and the theorem earns its name only when the transformations are actually produced. Notice also what the proof does not use: it never uses a rotation or a reflection, because a circle looks the same from every direction, and it never needs the circles to be in any special position.

A second misconception: any dilation will do. When r₂ ≠ r₁, the dilation must be centered at the (translated) center of the circle. (When the radii are equal the scale factor is 1, the dilation moves nothing wherever it is centered, and the translation alone does the job.) A dilation with scale factor other than 1 centered somewhere else still produces a circle of radius r₂, but centered at the wrong point, so its image would not be circle 2. The order also matters for the coordinates: translating first puts the center where it belongs, and then dilating about that point leaves the center fixed.

What similarity buys us. Because all circles are similar with scale factor r₂/r₁, every length measured on a circle scales by that same factor. The circumference of circle 2 is (r₂/r₁) times the circumference of circle 1, so the ratio circumference ÷ radius is the same for every circle; that constant is 2π, and it is the reason a single number π works for every circle in the world. Areas scale by the square of the factor: circle 2 has (r₂/r₁)² times the area of circle 1. The same reasoning, applied to arcs cut off by equal central angles, is what makes radian measure possible later in this unit.

Worked proof, written in full: Given circle 1 with center A and radius r₁ and circle 2 with center B and radius r₂, prove they are similar. Let T be the translation sending A to B. T preserves distance, so T maps the set of points at distance r₁ from A onto the set of points at distance r₁ from B: the circle centered at B with radius r₁. Let D be the dilation centered at B with scale factor k = r₂/r₁ > 0. D sends each point at distance r₁ from B to the point on the same ray at distance k·r₁ = r₂ from B, and every point at distance r₂ from B arises this way. So D maps the circle centered at B with radius r₁ onto circle 2. The composition D∘T is a rigid motion followed by a dilation and maps circle 1 onto circle 2, so the circles are similar.

Worked coordinate example: Circle 1 has center (−4, 1) and radius 2; circle 2 has center (3, −2) and radius 5. Translation: (x, y) → (x + 7, y − 3), since 3 − (−4) = 7 and −2 − 1 = −3. Scale factor: 5/2. Dilation about (3, −2): (x, y) → (3 + (5/2)(x − 3), −2 + (5/2)(y + 2)). Test the point (−2, 1) of circle 1 (distance 2 from (−4, 1)): translation gives (5, −2); dilation gives (3 + (5/2)(2), −2 + 0) = (8, −2), whose distance from (3, −2) is 5. The point lands on circle 2.

Worked ratio consequence: Circle 1 has radius 4 and circle 2 has radius 6, so the scale factor is 3/2. The circumference of circle 1 is 2π·4 = 8π ≈ 25.13 and that of circle 2 is 2π·6 = 12π ≈ 37.70, and 12π ÷ 8π = 3/2, the scale factor. The areas are 16π ≈ 50.27 and 36π ≈ 113.10, with ratio 36/16 = 9/4 = (3/2)², the square of the scale factor. Lengths scale by k; areas scale by k².

Worked misconception: A student maps circle 1 (center (2, 3), radius 4) to circle 2 (center (−1, 5), radius 6) by dilating about the origin with factor 3/2 and then translating. The dilation sends the center (2, 3) to (3, 4.5) and the radius to 6; a translation by (−4, 0.5) then moves (3, 4.5) to (−1, 5). So this order can also work, but only because the student recomputed the translation after the dilation. Dilating about the origin and then translating by the original vector (−3, 2) would leave the center at (0, 6.5), not (−1, 5). The safest sequence is the one in the proof: translate center to center first, then dilate about that shared center.

Transformation record: For each pair of circles, write the translation as a rule (x, y) → (x + a, y + b), the scale factor k, and the center of the dilation, then verify one point: (i) center (0, 0) radius 3 to center (4, −1) radius 9; (ii) center (5, 5) radius 10 to center (5, −3) radius 4; (iii) center (−2, −2) radius 1 to center (−2, −2) radius 7. For (iii) say which of the two transformations is unnecessary and why.

Independent performance: Write a complete proof that all circles are similar, using letters A, B, r₁ and r₂, and then illustrate it with two circles of your own choosing whose centers and radii are all different. Success criteria: the proof names the translation and states why it maps circle 1 onto the circle of radius r₁ centered at B; it names the dilation with its center and scale factor and states why the image is exactly circle 2 (both directions of "a point is on the image if and only if it is on circle 2"); it ends by saying why a rigid motion followed by a dilation establishes similarity; and the illustration tracks two specific points through both transformations with the distance check shown for each.