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Math I · Mathematics
Transformations and similarity · Rotations

Rotations

Before you start, make sure you know: Angle classification (Math 6)

Picture a Ferris wheel at the state fair. It is 60 feet across, and we put its center at the origin of a coordinate grid. You hop into the car at the 3 o'clock position, and the ride turns a quarter turn counterclockwise. Where is your car now? Rotations let you answer questions like this with a few numbers.

What is a rotation?

A rotation turns the whole plane around one fixed point, like a spinning record. Two things define it: where the center is and how far you turn.

By convention, a positive angle means counterclockwise and a negative angle means clockwise. So a rotation of $-90^\circ$ turns the figure the same way a clock's hands move.

Turning by $\theta$ and by $\theta - 360^\circ$ ends in the same place, because a full turn changes nothing. For example, $+270^\circ$ (three quarter turns counterclockwise) gives the same result as $-90^\circ$ (one quarter turn clockwise).

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A graphic designer is creating a pattern and applies a rotation of $270^\circ$ in the positive direction to a figure. His colleague says that the same result would have…

A graphic designer is creating a pattern and applies a rotation of $270^\circ$ in the positive direction to a figure. His colleague says that the same result would have been obtained by applying a single rotation in the negative (clockwise) direction. What angle should that negative rotation be for the figures to coincide?

Solve

Rotating about the origin

Let's rotate the point $P(4, 2)$ by $90^\circ$ counterclockwise about the origin $O$. Think of $P$ as "4 to the right, 2 up." After a quarter turn counterclockwise, "right" becomes "up" and "up" becomes "left." So $P'$ is "4 up, 2 left," which is $(-2, 4)$.

That works for any point, and it gives three rules to remember:

Rotation about $O(0, 0)$ $P(x, y) \to$ Example with $P(4, 2)$
$90^\circ$ $(-y,\ x)$ $(-2, 4)$
$180^\circ$ $(-x,\ -y)$ $(-4, -2)$
$270^\circ$ (or $-90^\circ$) $(y,\ -x)$ $(2, -4)$
The point P(4, 2) and its images after counterclockwise rotations of 90, 180 and 270 degrees about the origin, all on the same circle. P(4, 2) 90°: (−2, 4) 180°: (−4, −2) 270°: (2, −4) O 90°: (−y, x) 180°: (−x, −y) 270°: (y, −x)

All four points sit on the same circle centered at $O$, which is the "same distance from the center" part of the definition.

Drag the point below and check the rule against what you see.

To rotate a polygon, rotate each vertex and then connect the images in the same order.

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Given point $P(4, 2)$ on the Cartesian plane, perform a rotation with center at the origin of coordinates $O(0, 0)$ and angle $\alpha = 90^\circ$ in the positive…

Given point $P(4, 2)$ on the Cartesian plane, perform a rotation with center at the origin of coordinates $O(0, 0)$ and angle $\alpha = 90^\circ$ in the positive (counterclockwise) direction. What are the coordinates of the transformed point $P'$?

x y 4 2 −2 4 α = 90° O(0,0) P(4,2) P'(−2,4)
Solve
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A graphic designer is creating a new logo based on a triangle with vertices $A(2, 1)$, $B(5, 1)$ and $C(4, 3)$. To complete the design, they need to rotate this triangle…

A graphic designer is creating a new logo based on a triangle with vertices $A(2, 1)$, $B(5, 1)$ and $C(4, 3)$. To complete the design, they need to rotate this triangle with its center at the origin $O(0, 0)$ and an angle of $+90^\circ$ (counterclockwise). Graphically represent the original triangle and its image $A'B'C'$, stating the coordinates of the new vertices.

X Y +90° O(0,0) A(2,1) B(5,1) C(4,3) A'(−1,2) B'(−1,5) C'(−3,4)
Solve

Rotating about a point other than the origin

If the center is $C(h, k)$, the origin rules don't work directly. The trick is to rotate the vector from the center to the point, then walk back from the center.

Why does this work? Rotating about $C$ only cares about where $P$ is relative to $C$. Step 1 moves $C$ to the origin, step 2 does the rotation, and step 3 moves everything back.

Point P(5, 3) is rotated 90 degrees counterclockwise about C(2, 1): the step 3 right and 2 up becomes 3 up and 2 left, giving P'(0, 4). C P P′ 3 2 3 2
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Consider a square centered at point $C(2, 2)$ with sides parallel to the coordinate axes and of length $2$ units. a) State the coordinates of its four vertices. b) If we…

Consider a square centered at point $C(2, 2)$ with sides parallel to the coordinate axes and of length $2$ units.

  1. a) State the coordinates of its four vertices.
  2. b) If we apply a rotation of $+90^\circ$ with center at the square's own center $C(2, 2)$, what are the new coordinates of the vertices?
  3. c) Are there any points of the figure that have not changed position after the rotation? Justify your answer.
x y 0 1 2 3 1 2 3 V₁ V₂ V₃ V₄ C(2,2) +90°
Solve
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A video game designer is programming the movement of a magic key on a Cartesian plane. The key is defined by a line segment joining the points $A(1, 1)$ and $B(1, 4)$.…

A video game designer is programming the movement of a magic key on a Cartesian plane. The key is defined by a line segment joining the points $A(1, 1)$ and $B(1, 4)$. The program must perform a rotation with center $C(3, 1)$ and angle $\alpha = 90^\circ$ counterclockwise.

Draw the coordinate system in your notebook, plot the points and the original figure, and determine the coordinates of the new points $A'$ and $B'$ after the rotation. Explain the procedure you have followed using a compass or by reasoning about the distances to the center of rotation.

x y 0 α = 90° A(1,1) B(1,4) C(3,1)
Solve
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A traffic radar located at point $R(2, 2)$ rotates sweeping a circular sector. An auxiliary camera located at the origin $O(0, 0)$ captures an image of a vehicle at…

A traffic radar located at point $R(2, 2)$ rotates sweeping a circular sector. An auxiliary camera located at the origin $O(0, 0)$ captures an image of a vehicle at position $V(5, 6)$.

If the radar rotates the vehicle by an angle of $270^\circ$ counterclockwise with respect to its own position $R$, determine the new coordinates $V'$ of the vehicle on the control center map. Use a square grid to perform the rotation step by step, analysing the horizontal and vertical displacements with respect to the center $R$.

X Y 270° +3 +4 −3 +4 O(0,0) R(2,2) V(5,6) V'(6,-1)
Solve

Angles that are not multiples of 90°

For angles like $45^\circ$ or $30^\circ$, you can't just swap coordinates. You have three options.

Compass and protractor. Draw the ray from $C$ through $P$. Place the protractor at $C$ and measure the angle $\theta$ in the right direction. Then use the compass, set to the length $CP$, to mark $P'$ on the new ray.

Reasoning with right triangles. A $45^\circ$ turn from an axis makes an isosceles right triangle, so both legs are equal.

The rotation formula (any angle, any center). Use the same three steps as before, but rotate the vector with sine and cosine.

Why does this work? For $\theta = 90^\circ$, $\cos\theta = 0$ and $\sin\theta = 1$, and the formula gives $(-b, a)$, the rule you already know. The formula is the same idea for every angle.

$\theta$ $\cos\theta$ $\sin\theta$
$30^\circ$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{1}{2}$
$45^\circ$ $\dfrac{\sqrt{2}}{2}$ $\dfrac{\sqrt{2}}{2}$
$60^\circ$ $\dfrac{1}{2}$ $\dfrac{\sqrt{3}}{2}$

To check any rotation, use the Pythagorean Theorem (or the distance formula) to confirm that the point is still the same distance from the center.

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In a laser security system, a mirror located at point $M(0, 2)$ must rotate a light beam. The beam originally arrives following the direction of the line segment joining…

In a laser security system, a mirror located at point $M(0, 2)$ must rotate a light beam. The beam originally arrives following the direction of the line segment joining $A(-4, 2)$ to $M$. If we apply a rotation with center $M$ and angle $\alpha = -45^\circ$ (clockwise), what will be the new position of the point that was at $A$?

Calculate the coordinates of the new point $A'$ approximately using technical drawing and check if the distance to the center of rotation remains constant using Pythagoras' theorem.

X Y 0 2.83 2.83 d = 4 d = 4 −45° M(0, 2) A(−4, 2) A'(−2.83, 4.83)
Solve

Half turns (point reflections)

A rotation of $180^\circ$ is called a half turn, or a point reflection. Some problems call it a central symmetry. Each point $P$ goes straight through the center to the other side, at the same distance. Turning $180^\circ$ clockwise or counterclockwise gives the same result.

So the center $C$ is the midpoint of $PP'$. With center $C(h, k)$:

$$P(x, y) \to P'(2h - x,\ 2k - y)$$

With center at the origin this becomes $P(x, y) \to P'(-x, -y)$.

A figure F and its image F' after a 180 degree rotation about the point O, which is the midpoint of every segment joining a point to its image. O F F′ (rotated 180°)
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We have a line segment with endpoints $A(1, 1)$ and $B(3, 1)$. If we apply a central symmetry with center at the origin $O(0, 0)$, we know that this transformation is…

We have a line segment with endpoints $A(1, 1)$ and $B(3, 1)$. If we apply a central symmetry with center at the origin $O(0, 0)$, we know that this transformation is equivalent to a rotation by a certain angle.

  1. What is that angle of rotation?

  2. Draw the original segment and its transformed segment $A'B'$.

X Y 1 1 180° A (1,1) B (3,1) A' (−1,-1) B' (−3,-1) O (0,0)
Solve
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In your notebook, draw a line segment $PQ$ with coordinates $P(1, 4)$ and $Q(3, 2)$. Apply a rotation to the segment with center $C(1, 1)$ and angle…

In your notebook, draw a line segment $PQ$ with coordinates $P(1, 4)$ and $Q(3, 2)$. Apply a rotation to the segment with center $C(1, 1)$ and angle $\alpha = 180^\circ$.

  1. a) What name is given to this special type of rotation when the angle is $180^\circ$?
  2. b) Find the coordinates of the transformed points $P'$ and $Q'$.
  3. c) Check by calculating distances whether the length of the segment has been preserved.
Solve

Finding the center of rotation

Suppose you only see a figure and its image. How do you find the center?

Think about one point $A$ and its image $A'$. The center $O$ must satisfy $OA = OA'$, so $O$ is equally far from $A$ and $A'$. The points equally far from two points form their perpendicular bisector. So the center lies on the perpendicular bisector of $AA'$. The same is true for every other pair $B, B'$.

Two points A and B and their images A' and B'. The perpendicular bisectors of AA' and BB' cross at the center of rotation O. A B A′ B′ O Blue: bisector of AA′ Orange: bisector of BB′
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We have two figures in the plane, $F$ and $F'$. We know that $F'$ is the figure obtained after applying a rotation to $F$. To find the center of rotation $O$, we have…

We have two figures in the plane, $F$ and $F'$. We know that $F'$ is the figure obtained after applying a rotation to $F$. To find the center of rotation $O$, we have joined point $A$ of the original figure with its corresponding point $A'$ in the rotated figure, and point $B$ with its corresponding point $B'$.

Explain in detail the geometric procedure you must follow using the perpendicular bisector of the segments $AA'$ and $BB'$ to locate the center of rotation. What would happen if the segments $AA'$ and $BB'$ were parallel?

F F′ A B C A′ B′ C′ O
Solve
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We have found a fragment of an ancient Arabic mosaic where a triangle with vertices $P(2, 2)$, $Q(5, 2)$ and $R(4, 4)$ has been displaced by an unknown rotation. We know…

We have found a fragment of an ancient Arabic mosaic where a triangle with vertices $P(2, 2)$, $Q(5, 2)$ and $R(4, 4)$ has been displaced by an unknown rotation. We know that the image of point $P$ after the movement is $P'(8, 2)$ and the image of $Q$ is $Q'(8, 5)$.

Use the properties of perpendicular bisectors to graphically find the center of rotation $O$ of this movement. Once located, state its coordinates and calculate the size of the angle of rotation $\alpha$ and its direction.

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 α = 90° P (2,2) Q (5,2) R (4,4) P' (8,2) Q' (8,5) R' (6,4) O (5,5)
Solve

Composing rotations

What happens if you rotate, and then rotate again?

Why do the angles add? Each rotation turns every line by its angle, and turning a line by $\alpha$ and then by $\beta$ turns it by $\alpha + \beta$.

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Two consecutive rotations have been applied to a triangle with the same center $O$: the first of $40^\circ$ and the second of $50^\circ$, both in the positive direction.…

Two consecutive rotations have been applied to a triangle with the same center $O$: the first of $40^\circ$ and the second of $50^\circ$, both in the positive direction.

  1. What single rotation is this composition equivalent to?

  2. If we now apply a third rotation so that the triangle returns to its original position, what angle and direction must that rotation be?

+40° +50° +90° ? T₀ T₁ T₂ O
Solve
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We perform a composition of two transformations on a point $P(3, -1)$. First we apply a rotation with center $O(0, 0)$ and angle $\alpha = +90^\circ$ obtaining $P'$.…

We perform a composition of two transformations on a point $P(3, -1)$. First we apply a rotation with center $O(0, 0)$ and angle $\alpha = +90^\circ$ obtaining $P'$. Next, we apply to $P'$ another rotation with the same center $O$ and angle $\beta = +180^\circ$ obtaining $P''$.

  1. a) Determine the final coordinates of point $P''$.
  2. b) Is there a single rotation that allows us to go directly from $P$ to $P''$? If so, state its center and angle.
  3. c) Would the same result be obtained if we applied the rotation of $180^\circ$ first and then that of $90^\circ$? Justify whether the composition of rotations with the same center satisfies the commutative property.
Solve
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Consider a square with vertices $A(2, 1)$, $B(4, 1)$, $C(4, 3)$ and $D(2, 3)$. A composition of movements is applied to it: first a rotation with center $O(0, 0)$ and…

Consider a square with vertices $A(2, 1)$, $B(4, 1)$, $C(4, 3)$ and $D(2, 3)$. A composition of movements is applied to it: first a rotation with center $O(0, 0)$ and angle $180^\circ$, and to the result, another rotation with center $C'(-4, -3)$ (which is the corresponding point of $C$ after the first rotation) and angle $90^\circ$ counterclockwise.

  1. Determine the coordinates of the vertices after the first movement.

  2. Determine the final coordinates after the composition.

  3. Is there a single rotation that produces the same result? If so, what would its angle be?

Solve

Two reflections make a rotation

Recall that a reflection over a line $\ell$ sends $P$ to the point $P'$ such that $\ell$ is the perpendicular bisector of $PP'$. A reflection flips orientation, but two flips cancel out, so the result of two reflections keeps orientation. That is exactly what a rotation does.

Lines r and s meet at O at 25 degrees. Point P is reflected over r to P' and then over s to P''. The angle P O P'' is 50 degrees, twice the angle between the lines. O P P′ P″ r s 25° 50°

Here $r$ and $s$ meet at $25^\circ$. The point $P$ is $70^\circ$ above $r$. Reflecting over $r$ puts $P'$ at $-70^\circ$. Reflecting over $s$ then puts $P''$ at $2 \cdot 25^\circ - (-70^\circ) = 120^\circ$. The angle from $P$ to $P''$ is $120^\circ - 70^\circ = 50^\circ = 2 \cdot 25^\circ$.

To check it on paper: reflect $P$ over $r$ by drawing a perpendicular to $r$ and measuring the same distance on the other side. Repeat with $s$. Then measure $\angle POP''$ with a protractor.

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Show graphically in your notebook that the composition of two reflections in two axes $r$ and $s$ that intersect at a point $O$ forming an angle of $30^\circ$ is…

Show graphically in your notebook that the composition of two reflections in two axes $r$ and $s$ that intersect at a point $O$ forming an angle of $30^\circ$ is equivalent to a rotation.

To do this:

  1. Draw any point $P$ outside the axes.

  2. Find $P'$, its reflection in $r$.

  3. Find $P''$, the reflection of $P'$ in $s$.

  4. Measure the angle $\widehat{POP''}$. What is its relationship with the angle formed by the axes? State the general property that relates the angle of the axes to the angle of the resulting rotation.

Solve

Rotational symmetry

A figure has rotational symmetry if some rotation about a point, by less than a full turn, maps the figure exactly onto itself. The order of the rotational symmetry is the number of times the figure matches itself during one full turn (counting the full turn itself).

A fan with 5 identical blades has rotational symmetry of order 5, so the smallest angle is 72 degrees. 5 identical blades: order 5 smallest angle: 360° ÷ 5 = 72° A regular polygon with n sides has order n.

Why $\dfrac{360^\circ}{n}$? The $n$ matching positions are equally spaced around a full turn of $360^\circ$, so the gap between neighbors is $360^\circ \div n$.

Counting positions works the same way. A pizza cut into 8 equal slices has $360^\circ \div 8 = 45^\circ$ per slice. If you turn it $135^\circ$, each slice moves $135^\circ \div 45^\circ = 3$ slice positions. In general, positions moved $=$ angle $\div$ angle per slice.

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A ceiling fan has $5$ blades distributed completely symmetrically. If the fan is switched off and you push it gently causing one of the blades to move to the exact…

A ceiling fan has $5$ blades distributed completely symmetrically. If the fan is switched off and you push it gently causing one of the blades to move to the exact position where the next one was, what angle has the fan rotated? State the angle in both the positive (counterclockwise) and negative (clockwise) direction.

1 2 3 4 5 +? -? Counterclockwise Clockwise
Solve
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Look at the recycling logo made up of three arrows. What is the order of this center of rotation? From that order, calculate the minimum angle needed for the logo to…

Look at the recycling logo made up of three arrows. What is the order of this center of rotation? From that order, calculate the minimum angle needed for the logo to coincide exactly with itself when rotated.

1 2 3 α
Solve
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In a pizzeria, they have cut a family pizza into $8$ equal slices. If you rotate the circular tray by an angle of $135^\circ$ in a clockwise direction, how many slice…

In a pizzeria, they have cut a family pizza into $8$ equal slices. If you rotate the circular tray by an angle of $135^\circ$ in a clockwise direction, how many slice positions has each piece of pizza moved?

0 1 2 3 45° 45° 45° 135°
Solve

All the symmetry angles, and lines of symmetry

If the smallest symmetry angle is $\dfrac{360^\circ}{n}$, then every symmetry angle is a multiple of it: $\dfrac{360^\circ}{n},\ 2 \cdot \dfrac{360^\circ}{n},\ \dots$, up to just below $360^\circ$. Doing the smallest turn again and again produces all the others. That gives $n - 1$ rotations below $360^\circ$.

You can also work backward: a rotation of $\alpha$ maps a regular $n$-gon onto itself exactly when $\alpha$ is a multiple of $\dfrac{360^\circ}{n}$.

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In an Arabic mosaic, a minimum motif is used which is repeated by rotations. If the center of a regular polygon of $n$ sides is the center of rotation, and we know that…

In an Arabic mosaic, a minimum motif is used which is repeated by rotations. If the center of a regular polygon of $n$ sides is the center of rotation, and we know that when rotating the figure by an angle of $40^\circ$ it does not coincide with the original, but when rotating it $72^\circ$ it does:

  1. a) What is the minimum number of sides the regular polygon must have?
  2. b) Calculate all the angles of rotation less than $360^\circ$ that leave the figure invariant.
  3. c) If the polygon were a regular octagon, what would be the minimum angle of rotation for the figure to map onto itself?
40º 72º 45º
Solve
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A circular logo has a design that is repeated by rotations. If the design coincides with itself for the first time when rotated $40^\circ$ about its center: 1. What is…

A circular logo has a design that is repeated by rotations. If the design coincides with itself for the first time when rotated $40^\circ$ about its center:

  1. What is the order of rotational symmetry of this figure?

  2. State all the angles of rotation less than $360^\circ$ that leave the figure invariant.

  3. If the logo also had axes of symmetry, how many axes would it have at a minimum? Justify your answer based on the relationship between the order of rotational symmetry and the axes of symmetry in regular polygons and rosettes.

40º O
Solve

Rotation over time: clocks and turbines

Many real rotations happen at a constant speed, so the angle is proportional to the time. If one full turn takes $T$ seconds, then after $t$ seconds the rotation is

$$\frac{t}{T} \cdot 360^\circ$$

(use a negative sign if it turns clockwise).

On a clock, the minute hand turns $360^\circ$ in 60 minutes, so $6^\circ$ per minute. The hour hand turns $360^\circ$ in 12 hours (720 minutes), so $0.5^\circ$ per minute. Both turn clockwise, so their angles are negative.

When a point turns through an angle $\theta$ on a circle of radius $r$, the distance traveled is an arc of the circle:

$$\text{arc length} = \frac{|\theta|}{360^\circ} \cdot 2\pi r.$$
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Look at the following figure which represents a wind turbine blade. If the blade is a polygon with rotational symmetry, determine: a) The order of rotational symmetry of…

Look at the following figure which represents a wind turbine blade. If the blade is a polygon with rotational symmetry, determine:

  1. a) The order of rotational symmetry of the figure.
  2. b) The minimum angle it must rotate so that the blade coincides exactly with its initial position.
  3. c) If the blade rotates at a constant speed and takes $2$ seconds to complete a full rotation, what angle will it have rotated after $0,5$ seconds? Express the result in degrees and determine if at that moment the figure visually coincides with the original.
1 rotation = 2 s α
Solve
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A wall clock shows exactly $12:00$ o'clock. a) Calculate the angle of rotation (in degrees) described by the minute hand when $20$ minutes pass. Remember to indicate the…

A wall clock shows exactly $12:00$ o'clock.

  1. a) Calculate the angle of rotation (in degrees) described by the minute hand when $20$ minutes pass. Remember to indicate the sign of the angle according to the direction of rotation of the clock.
  2. b) If the minute hand measures $10\text{ cm}$, what is the distance traveled by the tip of the hand in those $20$ minutes? (Consider the path as an arc of a circle).
  3. c) What angle will the hour hand have rotated in that same time interval?
Solve

Fixed points and invariant lines

An invariant (fixed) point does not move. An invariant line is mapped onto itself, even though its points may slide along it.

For a rotation about $C$ by an angle $\alpha$ that is not a multiple of $360^\circ$:

  • The only fixed point is the center $C$. Any other point is at distance $r > 0$ from $C$, and it turns through a nonzero angle on its circle.

  • There are no invariant lines, with one exception: a rotation of $180^\circ$. A line through $C$ is turned by $\alpha$, so it lands back on itself only when $\alpha$ is a multiple of $180^\circ$. A $180^\circ$ turn sends every line through $C$ onto itself (with the two sides swapped). Lines that do not go through $C$ are sent to a parallel line on the other side of $C$.

Half turns about two different centers give a useful result. Rotate about $O_1(0, 0)$ by $180^\circ$ and then about $O_2(3, 0)$ by $180^\circ$. Take $P(1, 2)$. It goes to $(-1, -2)$, and then to $(2 \cdot 3 - (-1),\ 0 - (-2)) = (7, 2)$. Every point moves 6 units to the right, which is $2 \cdot O_1O_2$. In general, two half turns about different centers combine into a translation by twice the distance between the centers.

Now suppose a figure maps onto itself under half turns about two different centers. Then it maps onto itself under that translation too, and under it again and again. A point of the figure would then have infinitely many images, all in the figure. So such a figure must be infinite (a line, a strip, a repeating pattern).

Now put everything together on true-and-false statements and on the invariants of a rotation.

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State which of the following statements about rotations in the plane are true and which are false. Briefly justify your answer: a) In a rotation of $180^\circ$, the only…

State which of the following statements about rotations in the plane are true and which are false. Briefly justify your answer:

  • a) In a rotation of $180^\circ$, the only point that does not move is the center of rotation.

  • b) A rotation is an indirect isometry because it changes the orientation of angles.

  • c) If we rotate a circle around its own center by any angle, the resulting figure is identical to the original.

  • d) The composition of two rotations with a different center is always another rotation with the same center as the first.

Solve
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Analyse the invariant elements of a rotation in the plane. If we apply a rotation with center $C(x, y)$ and an angle $\alpha \neq 0^\circ$ and $\alpha \neq 360^\circ$:…

Analyse the invariant elements of a rotation in the plane. If we apply a rotation with center $C(x, y)$ and an angle $\alpha \neq 0^\circ$ and $\alpha \neq 360^\circ$:

  1. How many invariant points (that do not change position) exist in the whole plane?

  2. Are there invariant lines (that are transformed into themselves even if their points change position)? Study the special case where the angle of rotation is $180^\circ$.

  3. If a figure has two different centers of point reflection, $O_1$ and $O_2$, show that the figure must be infinite. (Hint: consider the composition of two rotations of $180^\circ$ and what movement results).

Solve

Back to the start: the Ferris wheel has radius 30 feet, so your car at the 3 o'clock position is the point $(30, 0)$. A quarter turn counterclockwise is a $90^\circ$ rotation about the origin, and $(x, y) \to (-y, x)$ gives $(0, 30)$: the very top, 30 feet above the center.

Summary

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