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Circles, Chords, and Arcs: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Compare three circles

Circle A has center O and radius 7 cm. Circle B has center P and radius 7 cm, with P different from O. Circle C has center O and radius 12 cm.

Describe the relationship between A and B.

Describe the relationship between A and C, and give a similarity mapping A onto C.
1. Hint 1
  Compare center locations separately from radii.
2. Hint 2
  A rigid motion preserves radius; a dilation can change it.
3. Hint 3
  For A to C, divide the target radius by the original radius.
Answer

A and B have different centers.
$A\cong B.$
A and C are concentric.
$\begin{aligned}A&\sim C,\\k&=\frac{12}{7}.\end{aligned}$
The dilation is about O.

Full solution

A and B have equal radii, so translation carrying O to P maps circle A onto circle B.
They are congruent.
Since O and P differ, they are not concentric.

A and C share O, so they are concentric.
Their radii differ, so they are not congruent.

Dilate about O by $\frac{12}{7}$ .
Every point 7 cm from O goes to distance 12 cm, and the inverse factor $\frac{7}{12}$ returns every point of C to A.
Thus the mapping covers the whole circle C.

Answer

A and B have different centers.
$A\cong B.$
A and C are concentric.
$\begin{aligned}A&\sim C,\\k&=\frac{12}{7}.\end{aligned}$
The dilation is about O.

Key idea

A shared center and an equal radius answer different circle-comparison questions.
Problem 2 Three routes around a circle

Points $A,B,C,D$ occur in that order around a circle. Consecutive minor arcs $\widehat{AB}$ , $\widehat{BC}$ , and $\widehat{CD}$ measure $54^\circ$ , $67^\circ$ , and $108^\circ$ .

Find $m(\widehat{DA})$ for the remaining consecutive arc.

Find the minor arc $\widehat{AC}$ .

Find the major arc $\widehat{ADC}$ .

Consecutive arcs follow A, B, C, D and return to A. Schematic.

Text description of this figure

Points A, B, C, D lie on a circle with center O. Arc AB is labeled 54°. Arc BC is labeled 67°. Arc CD is labeled 108°. Schematic; arc order and incidence are as shown.
1. Hint 1
  The four consecutive arcs together make one full turn.
2. Hint 2
  For the route from A through B to C, add only AB and BC.
3. Hint 3
  The two routes between A and C total 360 degrees.
Answer

$\begin{aligned}m(\widehat{DA})&=131^\circ,\\m(\widehat{AC})&=121^\circ,\\m(\widehat{ADC})&=239^\circ.\end{aligned}$

Full solution

The given arcs total 229 degrees.
$54^\circ+67^\circ+108^\circ=229^\circ.$
The remaining arc completes the circle.
$m(\widehat{DA})=360^\circ-229^\circ.$
Its measure is 131 degrees.

The route through B has measure
$m(\widehat{AC})=54^\circ+67^\circ.$
This is 121 degrees, so it is the minor arc.

The route through D is the complement of that minor arc within the full circle.
$m(\widehat{ADC})=360^\circ-121^\circ.$
Its measure is 239 degrees, which is greater than 180 degrees.

Answer

$\begin{aligned}m(\widehat{DA})&=131^\circ,\\m(\widehat{AC})&=121^\circ,\\m(\widehat{ADC})&=239^\circ.\end{aligned}$

Key idea

Track the selected route and the total turn before adding or subtracting arc measures.
Problem 3 An arc expression from a chord match

Chords $AB$ and $CD$ are congruent in one circle. The minor arc $\widehat{AB}$ measures $88^\circ$ . The minor arc $\widehat{CD}$ has measure $(5x-7)^\circ$ .

Find x.

Find the measure of the major arc from $C$ to $D$ .

The chords share one circle. The expressions for their arcs are stated in the problem.

Text description of this figure

Two equal chords AB and CD of one circle with center O. Radii join the center to all endpoints. The lower chord is dashed. Schematic.
1. Hint 1
  Identify the condition that allows equal chords to give equal minor arc measures.
2. Hint 2
  Set the expression for the CD minor arc equal to 88.
3. Hint 3
  The major arc is what remains in the full turn.
Answer

$\begin{aligned}x&=19,\\m(\text{major arc }CD)&=272^\circ.\end{aligned}$

Full solution

The chords belong to the same circle, so equal lengths give equal minor arc measures.
$\begin{aligned}5x-7&=88,\\5x&=95,\\x&=19.\end{aligned}$

The minor arc CD therefore measures 88 degrees.
Subtract it from 360 degrees to find the major arc.
$360^\circ-88^\circ=272^\circ.$

Answer

$\begin{aligned}x&=19,\\m(\text{major arc }CD)&=272^\circ.\end{aligned}$

Key idea

The equal-radius hypothesis is what connects chord equality to arc equality.
Problem 4 Test a missing hypothesis

Consider this statement: equal chord lengths in any two circles force equal perpendicular distances from their centers.

Give a numerical counterexample using two circles of different radii. Specify each radius, the common chord length, and the two perpendicular center distances. Verify that your chords exist.
1. Hint 1
  A counterexample needs equal chord lengths but unequal center distances.
2. Hint 2
  Keep one positive half chord fixed and choose two different positive perpendicular distances.
3. Hint 3
  Use Pythagoras to choose a radius for each resulting right triangle.
Answer

Common chord length 14 units.
$\begin{aligned}r_1&=\sqrt{65},\\d_1&=4,\\r_2&=\sqrt{85},\\d_2&=6.\end{aligned}$

Full solution

Choose half chord 7 units in both circles.
Choose perpendicular center distances 4 units and 6 units.

The first radius is chosen from the right triangle.
$\begin{aligned}r_1^2&=7^2+4^2,\\r_1&=\sqrt{65}.\end{aligned}$

The second radius is chosen in the same way.
$\begin{aligned}r_2^2&=7^2+6^2,\\r_2&=\sqrt{85}.\end{aligned}$

Each radius is greater than its chosen distance, so each line cuts off a genuine chord of length 14.
The two equal chord lengths have different center distances, 4 and 6.
The equal-radius hypothesis cannot be dropped.
Other verified numerical counterexamples are also valid.

Two actual radius-distance-half-chord triples. Both chords are 14 units long. Schematic.

Answer

Common chord length 14 units.
$\begin{aligned}r_1&=\sqrt{65},\\d_1&=4,\\r_2&=\sqrt{85},\\d_2&=6.\end{aligned}$

Key idea

A valid counterexample satisfies the hypothesis while breaking the claimed conclusion.
Problem 5 A brace across a circular frame

A circular window frame has radius 18 cm. A straight brace joins two points of the frame, and the perpendicular distance from the center to the brace is 9 cm. Find the brace length exactly.

The brace is a chord of the circular frame. Its length remains unknown. Schematic.

Text description of this figure

Chord AB has perpendicular foot M from center O. Radius OA is labeled 18 cm, perpendicular distance OM is labeled 9 cm, and half chord MB is labeled h. Schematic, not to scale.
1. Hint 1
  Join the center to one brace endpoint and use the perpendicular foot.
2. Hint 2
  The perpendicular through the center splits the brace into equal halves.
3. Hint 3
  Use the radius as the hypotenuse, then double the positive half length.
Answer

$18\sqrt3$ cm.

Full solution

Let h cm be half the brace length.
The radius, the perpendicular center distance and h form a right triangle.

Pythagoras gives
$\begin{aligned}h^2&=18^2-9^2,\\h^2&=243,\\h&=9\sqrt3.\end{aligned}$

The brace has two equal halves, so its length is
$2(9\sqrt3)=18\sqrt3\text{ cm}.$

Answer

$18\sqrt3$ cm.

Key idea

A physical bar between two points on a circular frame can be modeled as a chord.
Problem 6 A diameter divides both arcs

In a circle with center $O$ , chord $AB$ is not a diameter. The perpendicular from $O$ meets $AB$ at $M$ . The line $OM$ meets the minor arc $AB$ at $U$ and the major arc $AB$ at $V$ .

Prove that minor arcs $\widehat{AU}$ and $\widehat{UB}$ are congruent.

Prove that minor arcs $\widehat{AV}$ and $\widehat{VB}$ are congruent.

U selects the minor route and V selects the major route. Each requested pair lies in this one circle.

Text description of this figure

Chord AB is not a diameter. Perpendicular OM meets it at M. The diameter line OM meets the minor arc AB at U and the major arc AB at V. Endpoint radii OA and OB are drawn.
1. Hint 1
  Compare right triangles OMA and OMB.
2. Hint 2
  Equal central angles give equal arc measures in the same circle.
3. Hint 3
  Rays OU and OV are opposite, so each angle to OV supplements the matching angle to OU.
Answer

$\begin{aligned}\widehat{AU}&\cong\widehat{UB},\\\widehat{AV}&\cong\widehat{VB}.\end{aligned}$

Full solution

The chord is not a diameter, so O and M are distinct.
The endpoint radii are equal, OM is common, and the angles at M are right.
HL gives congruence of triangles OMA and OMB.

Their angles at O are equal.
Ray OU is ray OM because U lies on the minor arc beyond M.
Therefore
$m\angle AOU=m\angle UOB.$
The arcs AU and UB have equal degree measures.
They are minor arcs in the same circle, so they are congruent.

For V, ray OV is opposite ray OU.
Thus
$m\angle AOV=180^\circ-m\angle AOU.$
Also
$m\angle VOB=180^\circ-m\angle UOB.$
Subtracting equal angles from 180 degrees gives equal results.
The minor arcs AV and VB therefore have equal measures in the same circle and are congruent.

Answer

$\begin{aligned}\widehat{AU}&\cong\widehat{UB},\\\widehat{AV}&\cong\widehat{VB}.\end{aligned}$

Key idea

A center perpendicular determines matching central angles on both sides of the chord.
Problem 7 The zero-distance boundary

A circle has radius 19 cm. Chord AB has perpendicular center distance zero. Chord CD has the same perpendicular center distance as AB.

Find the length of AB.

Find the length of CD and identify the common midpoint of both chords.

A diameter passes through its center. The perpendicular distance is zero.

Text description of this figure

Diameter AB passes through center O. A perpendicular line through O meets it at its midpoint. Radius is 19 cm.
1. Hint 1
  Distance zero means the center lies on the chord line.
2. Hint 2
  A chord through the center is a diameter.
3. Hint 3
  The equal-distance result includes two diameters, even though no nondegenerate center-to-chord right triangle exists.
Answer

$\begin{aligned}AB&=38\text{ cm},\\CD&=38\text{ cm}.\end{aligned}$
Both midpoints are the center.

Full solution

Since AB has center distance zero, its line contains the center.
The chord AB is a diameter.

A diameter has two radii, so
$\begin{aligned}AB&=2(19),\\AB&=38\text{ cm}.\end{aligned}$

CD also has center distance zero, so CD is another diameter of the same circle.
Its length is 38 cm, and each diameter has the center as its midpoint.
No triangle with a zero-length leg is needed.

Answer

$\begin{aligned}AB&=38\text{ cm},\\CD&=38\text{ cm}.\end{aligned}$
Both midpoints are the center.

Key idea

The diameter case is handled directly from the center and the two opposite radii.
Problem 8 One line in concentric circles

Two concentric circles have center O and radii 20 cm and 37 cm. A line cuts a 32 cm chord AB from the smaller circle and a chord CD from the larger circle.

Find the perpendicular distance from O to the line.

Find CD.

The center and chord line are shared; the radii differ. Schematic.

Text description of this figure

Concentric circles share center O. One line contains inner chord AB and outer chord CD. A perpendicular from O meets both chords at the same midpoint. Schematic, not to scale.
1. Hint 1
  The perpendicular foot from the shared center is the midpoint of both chords.
2. Hint 2
  First use half of the known smaller chord and the smaller radius.
3. Hint 3
  Use the same perpendicular distance with the larger radius to find a new half chord.
Answer

$\begin{aligned}d&=12\text{ cm},\\CD&=70\text{ cm}.\end{aligned}$

Full solution

The smaller chord has half length 16 cm.
The perpendicular from O bisects it.

Use the smaller right triangle.
$\begin{aligned}d^2&=20^2-16^2,\\d^2&=144,\\d&=12.\end{aligned}$

The same line has the same distance from the same center.
Its larger half chord h uses radius 37 cm.

Pythagoras gives
$\begin{aligned}h^2&=37^2-12^2,\\h^2&=1225,\\h&=35.\end{aligned}$
Thus the full chord CD is 70 cm.

Answer

$\begin{aligned}d&=12\text{ cm},\\CD&=70\text{ cm}.\end{aligned}$

Key idea

Concentric circles share the center-to-line distance, while each radius determines its own chord length.
Problem 9 Parallel chords on opposite sides

A circle has radius $\sqrt{145}$ cm. Parallel chords $AB$ and $CD$ lie on opposite sides of the center. Their lengths are 18 cm and 16 cm. Find the perpendicular distance between the lines containing the chords.

The chord lines are parallel and the center lies between them. Schematic.

Text description of this figure

Parallel chords AB and CD lie on opposite sides of center O. The common perpendicular through O measures their separation. Schematic, not to scale.
1. Hint 1
  Find each perpendicular distance from the center separately.
2. Hint 2
  The half chords are 9 cm and 8 cm.
3. Hint 3
  Because the center lies between the parallel lines, combine the two distances by addition.
Answer

$17$ cm.

Full solution

Let the perpendicular center distances to AB and CD be p cm and q cm, respectively.
The chord halves are 9 cm and 8 cm.

For AB,
$\begin{aligned}p^2&=145-9^2,\\p&=8.\end{aligned}$

For CD,
$\begin{aligned}q^2&=145-8^2,\\q&=9.\end{aligned}$

The perpendicular through the center meets both parallel lines.
The center is between them, so the separation is the sum of the two distances.

The separation is
$8+9=17\text{ cm}.$

Answer

$17$ cm.

Key idea

The location of the center decides whether parallel-chord distances are added or subtracted.
Problem 10 Transfer through congruence and similarity

Circle O has radius $\sqrt{193}$ cm and chord AB of length 24 cm. A congruent circle P has chord CD whose minor arc has the same degree measure as minor arc AB.

Find CD and its perpendicular distance from P.

A dilation of factor 2 maps circle P onto circle Q and maps chord CD onto chord EF. Find EF and its perpendicular distance from Q.

The original circle has the stated 24 cm chord; h and d remain unknown. The other circles are specified by the mappings. Schematic.

Text description of this figure

Chord AB has perpendicular foot M from center O. Radius OA is labeled √193 cm, perpendicular distance OM is labeled d, and half chord MB is labeled h. Schematic, not to scale.
1. Hint 1
  The first transfer uses equal minor arcs in congruent circles.
2. Hint 2
  Compute the center distance from the original radius and half chord.
3. Hint 3
  A positive dilation scales all lengths, including a perpendicular center distance.
Answer

$\begin{aligned}CD&=24\text{ cm},\\d_P&=7\text{ cm},\\EF&=48\text{ cm},\\d_Q&=14\text{ cm}.\end{aligned}$

Full solution

Equal minor arc measures in congruent circles give equal chords.
Thus CD has length 24 cm.
The equal-radius hypothesis is supplied by congruence.

Half of AB is 12 cm.
Its perpendicular center distance d satisfies
$\begin{aligned}d^2&=193-12^2,\\d^2&=49,\\d&=7.\end{aligned}$

Equal chords in congruent circles have equal perpendicular center distances.
Hence the distance from P to CD is 7 cm.

The factor-2 dilation takes the entire circle and its chord to their images, multiplying lengths by 2.
Therefore
$\begin{aligned}EF&=2(24),\\EF&=48\text{ cm}.\end{aligned}$

A dilation preserves perpendicularity, so the image of the perpendicular from P is the perpendicular from Q.
Its length is
$\begin{aligned}d_Q&=2(7),\\d_Q&=14\text{ cm}.\end{aligned}$
The new radius is twice the original, and the minor arc degree measure is preserved.

Answer

$\begin{aligned}CD&=24\text{ cm},\\d_P&=7\text{ cm},\\EF&=48\text{ cm},\\d_Q&=14\text{ cm}.\end{aligned}$

Key idea

Use congruence for equal lengths and similarity for a known scale factor, carrying perpendicular distances with them.

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