Circles and Construction Tools

A circle records one fixed distance

For the distance classifications below, work in one chosen plane: the center and every point being classified lie in that plane.

Imagine a point $O$ and a length of $5$ centimeters. A point $P$ belongs to the circle exactly when $OP = 5$ centimeters. A point $3$ centimeters from $O$ is inside; one $7$ centimeters away is outside. Neither belongs to the circle itself.

A circle is the set of all points in a plane at one fixed positive distance from a point in that plane called its center. The fixed distance is the radius. For a circle with center $O$ and radius $r > 0$, a point $P$ in the circle’s plane belongs to the circle exactly when $OP = r$. The center is not on the circle because its distance from itself is zero.

A disk with center $O$ and positive radius $r$ consists of all points $P$ in that plane with $OP \le r$. It includes the circle, where $OP=r$, and the points of the plane with $OP<r$, called its interior.

Name the segments in a circle

The word radius can also mean a segment from the center to a point on the circle. Its length is the circle’s radius. Every such segment has the same length because that is the distance condition defining the circle.

A chord is a segment whose two endpoints lie on the circle. A diameter is a chord that passes through the center. Every diameter is a chord; a chord that misses the center is not a diameter.

If a circle has radius $5$ cm, a diameter $AB$ passes through $O$. The center is not an endpoint because $AO = BO = 5 > 0$. Since $O$ lies on the segment and is neither endpoint, it is between $A$ and $B$. The diameter therefore joins two radii end to end:

$AB = AO + OB = 5 + 5 = 10 \text{ cm}.$

In general, a diameter passes through the center, and neither endpoint is the center because each is a positive distance $r$ away. The center is therefore between the endpoints. Segment addition gives $d = r + r$, so a diameter’s length is twice the radius:

$d = 2r.$

This does not say that every chord has length $2r$. Passing through the center is the condition that lets us add those two radii along the chord.

What the construction tools may do

A compass fixes a distance between its point and pencil. Keeping its point at a center and its opening unchanged draws a circle, or an arc, a portion of a circle. We use a compass that can retain its opening when lifted and moved.

An unmarked straightedge draws the straight line through two distinct points. It can join points or extend a drawn segment. It has no measurement scale, so reading a length from numbered marks is not a straightedge construction.

Our ideal tools draw exact circles and lines. Real pencil marks have thickness, and a compass may slip. Measurement can check your paper work, but the construction’s reason explains why the ideal lengths agree.

All the constructions below take place in one chosen plane. We accept that a circle can be drawn in that plane with a given center in the plane and any radius $r > 0$. Every ray in that same plane starting at the center meets the circle exactly once.

Specific intersection facts for the circle pairs are stated at the construction step that uses them.

Copy a segment by keeping the opening

Start with a nonzero segment $AB$ and a ray $PX$ on which to place its copy. Set the compass point at $A$ and its pencil at $B$. The opening now equals $AB$. The first circle below records that setting.

Without changing the opening, place the compass point at $P$. Draw an arc or circle crossing ray $PX$ at $Q$. Then $\overline{PQ}$ is the required copy. For example, if the original length is $4$ cm, the new length is also $4$ cm even though no ruler reading was used to place $Q$.

Use Previous to return to the givens, then advance the stages. Before revealing $Q$, predict which distance the new circle fixes.

The finished stages keep both circles. Matching ticks record the copied lengths.

Copy an angle with two compass settings

For this procedure, use a selected opening greater than $0^\circ$ and less than $180^\circ$. Let the source angle have vertex $O$. Draw a target ray $PX$.

With any convenient positive compass opening, draw a circle centered at $O$. Let it meet the source rays at $A$ and $B$. Keep that opening and draw a circle centered at $P$; let it meet $PX$ at $D$. Thus $OA$, $OB$, and $PD$ have the same length.

Now change the compass opening to the distance $AB$, the chord between the source marks. With center $D$, draw a circle of that radius. In our chosen plane, we accept that this circle and the target circle centered at $P$ meet at two points on opposite sides of their center line $PD$, for the stated source opening strictly between $0^\circ$ and $180^\circ$ and positive initial radius. Since $D$ lies on ray $PX$, line $PD$ is the target baseline. Choose the intersection $E$ with the target circle on the desired side of $PX$. In these drawings the source opening is above its baseline, so choose the upper intersection at the target. Draw ray $PE$ to complete the copy.

For a numerical instance, suppose the first compass setting is $7$ cm and the distance $AB$ between the source marks is $5$ cm. Keep the $7$ cm setting to draw the target circle at $P$. Then change to $5$ cm for the circle centered at $D$. The resulting lengths are $OA = OB = PD = PE = 7$ cm and $DE = AB = 5$ cm. These are the transferred distance conditions; a protractor check of the copied opening is evidence about the paper drawing, rather than its exact justification.

Predict which setting is kept for the first two circles and which setting is changed for the third. Use Previous and Next to compare those actions.

The completed stages preserve both equal-radius circles and the chord-transfer circle. One tick marks the common radius; two ticks mark the copied chord length.

The completed procedure copies the source angle’s measure. A justification that the copied lengths force the angle to agree uses triangle congruence from chapter 4. Lesson 4.5, Using Congruence in Proofs, applies it to complete this argument. We defer that angle-copy argument until then. Checking with a protractor gives evidence about your drawing; it does not replace that proof.

Construct three equal sides

An equilateral triangle is a triangle whose three sides have equal lengths. Begin with distinct points $A$ and $B$. Draw a circle centered at $A$ with radius $AB$. Keep that opening and draw a second circle centered at $B$ in the same plane. For these distinct centers and the positive common radius $AB$, we accept that the two circles meet at two points on opposite sides of line $AB$. Choose either intersection as $C$, then join $AC$ and $BC$ with the straightedge. The existence of those points is accepted; the equal side lengths will follow from the circle definitions.

Before revealing the last two segments, predict what each circle tells you about a side. Use the stages to check your prediction against the retained circles.

The finished stages show the two distance conditions meeting at $C$. Choosing the lower intersection would produce another equilateral triangle on the other side of $AB$. The chosen $C$ is off line $AB$, so $A$, $B$, and $C$ are noncollinear.

The cumulative reference now gains these entries. The accepted facts supply the construction points; the proved results explain the equalities obtained from them.

Common mistakes