Circles, Chords, and Arcs

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Learning goals

Describe circle parts and distinguish concentric, congruent, and similar circles
Prove that translations and dilations map any positive-radius circle onto another
Calculate minor, major, and semicircle arc measures using central angles and addition
Justify chord and arc congruence with the same-circle or congruent-circle hypothesis
Apply perpendicular chord bisection and center distance, including the diameter case

One center, one fixed distance

A circle is the set of all points in a plane at one fixed positive distance from a fixed point, its center. A radius is a segment from the center to a point on the circle; the same word names its length. A chord joins two distinct points on the circle. A diameter is a chord through the center. It consists of two radii, so its length is twice the radius. The circle is the boundary; its interior is not part of that set of points.

A diameter is a chord through the center. Its midpoint is the center itself.

Two circles are concentric when they have the same center. Distinct concentric circles have different radii. Two circles are congruent when a rigid motion maps one onto the other. This happens exactly when their radii are equal: rigid motions preserve distance, and translating one center to the other then matches every point at that radius. Equal-radius circles may have different centers; congruent and concentric describe different relationships.

Why every circle is similar to every other

A change of size can carry an entire circle onto another circle.

A circle of radius $2$ can be translated to share the center of a circle of radius $7$ . A dilation by $\tfrac72$ about that center then sends every distance of $2$ to a distance of $7$ . Reversing that dilation sends every point of the second circle back to the first. Thus the match covers both whole circles, not just a few chosen points.

The center moves first. A positive dilation then adjusts every radius by the same factor.

Use the accepted transformation properties from Dilations and Scale Factors: a dilation about its center multiplies all distances from that center by its positive scale factor.

All positive-radius circles are similar#

For radius- $3$ and radius- $8$ circles, translate the first center to the second center. Every point on the translated circle is $3$ units from that center. Dilation by $\tfrac83$ sends it to distance $8$ . Conversely, every point at distance $8$ has one preimage on its ray at distance $3$ , obtained by the factor $\tfrac38$ .

Now take any circles with centers $O$ and $P$ and positive radii $r$ and $s$ . Translate $O$ to $P$ . Dilate about $P$ by the positive factor $s/r$ . Each point at distance $r$ goes to distance $s$ because

r\left(\frac{s}{r}\right)=s.

For any point of the target circle, the inverse dilation by $r/s$ produces a point at distance $r$ , and the inverse translation returns it to the original circle. These inverse steps also show that different points cannot merge. This is a one-to-one match onto the whole second circle, made from allowed similarity transformations. Nothing depended on the particular radii $3$ and $8$ .

Worked example 1 Map two whole circles

Problem. Circle $O$ has radius $4$ cm. Circle $P$ has radius $10$ cm. Describe a similarity carrying circle $O$ onto circle $P$ , and decide whether they are congruent.

The two centers and radii belong to this mapping problem. Schematic.

Solution. Translate $O$ to $P$ , then dilate about $P$ by the factor

\begin{aligned}k&=\frac{10}{4},\\k&=\frac52.\end{aligned}

Every original radius becomes $10$ cm. The inverse factor $\tfrac25$ takes every target point back, so the image is the whole target circle. The circles are similar. They are not congruent because their radii differ.

Check your understanding

Two distinct circles share a center, with radii $6$ cm and $9$ cm. Which description is correct?

A Concentric and congruent B Concentric and not congruent C Not concentric and congruent D Not concentric and not congruent

Arc measure records a turn

An arc is one of the two portions of a circle between distinct endpoints. A central angle has its vertex at the center and its sides along radii to those endpoints. Arc measure records the turn about the center, in degrees; it does not report the length along the curved boundary.

For example, if the smaller central angle from $OA$ to $OB$ is $80^\circ$ , the short arc has measure $80^\circ$ . The other way around measures $280^\circ$ , since the two turns complete $360^\circ$ .

The short arc from A to B and the arc from A through C to B use the same endpoints.

A minor arc measures less than $180^\circ$ . Two letters name it when the short route is unambiguous, as in $\widehat{AB}$ . A major arc measures more than $180^\circ$ and less than $360^\circ$ . Use three letters to select its route, as in $\widehat{ACB}$ , with $C$ on that major arc. If the endpoints are opposite ends of a diameter, each route is a semicircle, measuring $180^\circ$ . A third point selects which semicircle is meant; neither is a minor or major arc.

The degree measure of a minor arc is defined to be the measure of its smaller central angle. A major arc has the remaining measure in a full turn. These definitions use angle turns, so changing the radius does not change an arc’s degree measure. Equal degree measures on different-sized circles do not make the arcs congruent as figures.

Arcs that meet only at endpoints and trace a larger arc without overlap have measures that add. This follows from angle addition at the center. If $B$ lies on the selected arc from $A$ to $C$ , then

m(\widehat{ABC})=m(\widehat{AB})+m(\widehat{BC}).

Here the two smaller arcs are the routes through $B$ in that selected arc. Do not replace them by other routes between the same endpoints.

Worked example 2 Select the route before adding

Problem. Points $A,B,C,D$ occur in that order around a circle. The consecutive minor arcs $\widehat{AB}$ and $\widehat{BC}$ measure $74^\circ$ and $59^\circ$ . Find the minor arc $\widehat{AC}$ and the major arc $\widehat{ADC}$ .

The route from A through B to C is the sum of the two given adjacent arcs. Schematic.

Solution. Add the adjacent arcs through $B$ .

\begin{aligned}m(\widehat{AC})&=74^\circ+59^\circ,\\m(\widehat{AC})&=133^\circ.\end{aligned}

This is below $180^\circ$ , so it is the minor arc. The route through $D$ is the rest of the circle.

\begin{aligned}m(\widehat{ADC})&=360^\circ-133^\circ,\\m(\widehat{ADC})&=227^\circ.\end{aligned}

Check your understanding

Diameter $AC$ divides a circle into two semicircles. Point $B$ is on one of them, and $m(\widehat{AB})=112^\circ$ . Find the remaining arc $\widehat{BC}$ within that semicircle.

248^\circ

68^\circ

112^\circ

56^\circ

Equal chords and their arcs

In one circle, or in congruent circles, equal chord lengths go with equal minor arc measures. The reverse is also true. The equal-radius condition is essential.

Imagine two chords of length $12$ in circles of radius $8$ . Each chord and its two endpoint radii form a triangle with side lengths $8,8,12$ . SSS gives equal central angles, hence equal minor arc measures. Starting instead with equal central angles and two radius- $8$ circles, SAS gives equal chords.

The two pairs of radii let SSS or SAS connect chord lengths with central angles.

Equal chords and equal arcs in equal-radius circles#

Two $9,9,10$ triangles are congruent by SSS, so their angles opposite the $10$ -unit sides are equal. These are the smaller central angles. For the reverse direction, two $70^\circ$ central angles with adjacent radius lengths $9$ determine congruent triangles by SAS, so their chords are equal.

For the general case, let $AB$ and $CD$ be non-diameter chords in the same circle or in circles of the same positive radius. Join each center to both endpoints. If the chords have equal lengths, the two radius pairs and the chords give SSS congruence. The central angles, and therefore the minor arc measures, are equal.

Conversely, equal minor arc measures give equal central angles by the definition of arc measure. Together with the equal radii, these give SAS congruence and equal chord lengths. No step depended on the numbers $9$ , $10$ , or $70$ .

If a chord is a diameter, its endpoints give semicircles. Equal-length chords in equal-radius circles must then both be diameters, since no non-diameter chord reaches twice the radius (the triangle inequality is strict). Conversely, any two diameters of equal-radius circles have equal lengths. Their corresponding semicircles each measure $180^\circ$ .

Equal minor arcs also leave equal major arc measures after subtraction from $360^\circ$ . Equal major arc measures give equal minor arc measures in the reverse direction. In equal-radius circles, arcs of the same type with equal measure are congruent arcs: move the centers together, rotate one endpoint onto its partner, and reflect if needed to match the selected route. The entire selected arc matches. A minor and a major arc sharing one chord are not congruent to each other.

Worked example 3 Transfer chord equality to arc measures

Problem. In one circle, chords $AB$ and $CD$ are congruent. Their minor arcs have the following measures, in degrees. Find $x$ and the major arc from $C$ to $D$ .

\begin{aligned} m(\widehat{AB})&=(4x+6)^\circ,\\ m(\widehat{CD})&=(7x-24)^\circ. \end{aligned}

Both congruent chords lie in one circle; the given arc expressions are stated above.

Solution. The same-circle hypothesis lets us equate the minor arc measures.

\begin{aligned} 4x+6&=7x-24,\\ 30&=3x,\\ x&=10. \end{aligned}

Substitute into either expression, then subtract from a full turn.

m(\widehat{CD})=46^\circ.

The major arc from $C$ to $D$ therefore measures

360^\circ-46^\circ=314^\circ.

The $46^\circ$ result is between $0^\circ$ and $180^\circ$ , as a minor arc must be.

Check your understanding

Two congruent circles have equal minor arc measures at chords $JK$ and $LM$ . If $JK$ is $17$ cm long, what follows?

LM

34

cm long B

LM

8.5

cm long C

LM

17

cm long D The chord lengths cannot be compared

A perpendicular through the center

A line through the center of a circle that is perpendicular to a chord bisects that chord. For a non-diameter chord, the segment from the center to its midpoint is also perpendicular to the chord.

For instance, take a chord of length $10$ in a circle of radius $13$ . Its midpoint is $5$ units from each endpoint. The center is $13$ units from each endpoint. Both points lie on the chord’s perpendicular bisector by the perpendicular-bisector locus, so the line joining them is that perpendicular bisector.

For this non-diameter chord, the center and midpoint determine its perpendicular bisector.

Perpendicular chord bisection and its converse#

With radius $10$ and chord length $14$ , let $M$ be the midpoint. Both $O$ and $M$ are equally far from the chord endpoints, so both lie on the perpendicular bisector. They are distinct, and hence determine that line. Conversely, the perpendicular line through $O$ is that same unique perpendicular bisector and meets the chord at $M$ .

Now let $AB$ be any non-diameter chord with midpoint $M$ and center $O$ . The endpoint radii have equal lengths. Thus $O$ lies on the perpendicular bisector of $AB$ . The midpoint $M$ lies on it too. Since $O$ and $M$ are distinct, line $OM$ is that perpendicular bisector. Conversely, the unique perpendicular to line $AB$ through $O$ must be the perpendicular bisector already found, so its intersection with $AB$ is $M$ . The numeric lengths played no part beyond showing equal endpoint distances.

For a diameter $AB$ , its midpoint is $O$ because the two opposite radii have equal lengths. Every line through $O$ bisects that diameter, so a line through its midpoint need not be perpendicular. The segment $OM$ now has zero length and determines no line. The perpendicular line through $O$ still exists, and still bisects $AB$ at $O$ .

For a non-diameter chord, the two right triangles on either side of the perpendicular have equal hypotenuses (radii) and the common leg $OM$ . HL congruence shows that $OM$ also bisects the smaller central angle. The diameter line containing $OM$ therefore bisects the minor arc, and its opposite ray bisects the major arc. For example, a $110^\circ$ minor arc splits into two $55^\circ$ arcs; the $250^\circ$ major arc splits into two $125^\circ$ arcs. If $AB$ itself is a diameter, the perpendicular diameter makes four $90^\circ$ central angles, so it bisects both semicircles.

Worked example 4 Find a full chord from one right triangle

Problem. A circle has radius $13$ cm. The perpendicular distance from its center $O$ to chord $AB$ is $5$ cm. Find $AB$ .

The unknown h labels half the chord. The drawing is schematic, not a measuring tool.

Solution. Let $M$ be the perpendicular foot. Chord bisection gives two equal halves. In right triangle $OMA$ , the radius is the hypotenuse.

\begin{aligned} h^2+5^2&=13^2,\\ h^2&=144,\\ h&=12. \end{aligned}

A length is positive, so use the positive square root. The full chord contains both halves.

AB=2(12)=24\text{ cm}.

Check your understanding

A line through center $O$ meets a non-diameter chord $AB$ at its midpoint $M$ . Which justification establishes a right angle at $M$ ?

A Every line through a center is perpendicular to every chord B The midpoint must be on the circle C Equal radii make every triangle equilateral D The center and midpoint lie on the perpendicular bisector of

AB

Center distance compares chords

The distance from the center to a chord means the perpendicular distance to the line containing that chord. Its foot lies at the chord’s midpoint, so this is also the shortest distance to the chord itself.

A radius- $25$ circle with a chord $24$ units from its center gives a right triangle with hypotenuse $25$ and one leg $24$ . Its other leg is $7$ , since the squares are $625$ and $576$ , leaving $49$ . The full chord is $14$ units. This is the same Pythagorean reasoning used for any right triangle.

Chord length and perpendicular center distance#

For a radius- $17$ circle with center distance $8$ , half the chord satisfies

\begin{aligned} h^2&=17^2-8^2,\\ h^2&=225,\\ h&=15. \end{aligned}

Generally let the radius be $r$ , the perpendicular center distance be $d$ , and half the chord be $h$ . For a non-diameter chord, bisection and Pythagoras give

\begin{aligned} h^2+d^2&=r^2,\\ h&=\sqrt{r^2-d^2},\\ \text{chord length}&=2\sqrt{r^2-d^2}. \end{aligned}

The numeric argument used the radius and perpendicular distance only, so the same steps work for any positive radius and a non-diameter chord. For a diameter, $d$ is zero and the length is $2r$ directly from the definition. The formula agrees in this boundary case without pretending there is a nondegenerate right triangle.

A genuine chord has two distinct endpoints, so the permitted distances are

0\le d<r.

At distance $r$ , the formula would give zero, not a chord with distinct endpoints.

In the same circle or congruent circles, equal perpendicular center distances give equal values of $r^2-d^2$ and hence equal chord lengths. Conversely, equal chord lengths give equal half lengths; subtracting their equal squares from the same radius square gives equal distance squares. Distances are nonnegative, so the distances themselves are equal. This includes two diameters at distance zero.

With a fixed radius, increasing the nonnegative distance makes its square larger. That makes the positive half-chord square smaller and the chord shorter. A diameter, at distance zero, is therefore a longest chord.

A measurement or moving diagram can suggest this comparison. The right-triangle argument establishes it for every chord satisfying the hypotheses.

Try the four center distances. Predict how chord $AB$ changes, then compare its readout while the radius stays fixed.

A chord and its perpendicular center distance

Distance

The radius is 5 units. OH = 2 units. AB ≈ 9.17 units. The minor central angle AOB ≈ 132.84°.

Radius 5 stays fixed. OH is perpendicular to AB; the chord shortens as OH increases.

Worked example 5 Recover a distance, then transfer it

Problem. Two congruent circles have radius $15$ cm. In the first, chord $EF$ is $18$ cm long. In the second, chord $GH$ is also $18$ cm long. Find each chord’s perpendicular center distance and explain the transfer.

The full chord is 18 cm. Both the half chord h and the center distance d await calculation. Schematic.

Solution. Half of $EF$ is $9$ cm. If its center distance is $d$ cm, Pythagoras gives

\begin{aligned} d^2+9^2&=15^2,\\ d^2&=144,\\ d&=12. \end{aligned}

The chord $GH$ has the same length in a circle of the same radius. Equal chords in congruent circles have equal perpendicular center distances, so each distance is $12$ cm.

Check your understanding

In a circle of radius $13$ units, chord $AB$ is $4$ units from the center and chord $CD$ is $9$ units from the center. Which chord is longer?

AB

CD

C They are equal D Neither is a genuine chord

Circles, Chords, and Arcs

Learning goals

One center, one fixed distance

Why every circle is similar to every other

Arc measure records a turn

Equal chords and their arcs

A perpendicular through the center

Center distance compares chords

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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