What is Circles & arcs on the Geometry Regents?

Circles & arcs is a Circles topic on the Geometry Regents. RegentsLab has 21 practice questions on it, each with a full worked solution.

How do you study Circles & arcs for the Geometry Regents?

Work a range of Circles & arcs question types and read the worked solution every time you miss one — the exam reuses the same ideas year after year. Start with the questions on this page.

Circles & arcs: Geometry Regents Practice

Practice Circles & arcs for the Geometry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

Open the practice lab Score calculator

Question 25 · 2 pts

In circle $O$, the radius is $9$. A central angle of $80^\circ$ intercepts arc $AB$. What is the length of arc $AB$?

(A) $4\pi$ ✓
(B) $8\pi$
(C) $2\pi$
(D) $18\pi$

Worked solution

$\text{Arc length} = \frac{\theta}{360^\circ}\cdot 2\pi r$ Arc is a fraction of the full circumference

$= \frac{80}{360}\cdot 2\pi(9)$

$= \frac{2}{9}\cdot 18\pi = 4\pi$

Answer: $4\pi$

Arc length is the fraction $\frac{\theta}{360^\circ}$ of the full circumference $2\pi r$.

Question 26 · 2 pts

In circle $O$, inscribed angle $\angle ABC$ intercepts arc $AC$, which measures $128^\circ$. What is $m\angle ABC$?

(A) $64^\circ$ ✓
(B) $32^\circ$
(C) $128^\circ$
(D) $116^\circ$

Worked solution

$m\angle ABC = \tfrac{1}{2}\, m\widehat{AC}$ Inscribed angle theorem

$= \tfrac{1}{2}(128^\circ) = 64^\circ$

Answer: $64^\circ$

An inscribed angle equals half the measure of its intercepted arc.

Question 12 · 2 pts

In a circle, a central angle intercepts an arc of $80^\circ$. An inscribed angle intercepts the same arc. What is the measure of the inscribed angle?

(A) $160^\circ$
(B) $100^\circ$
(C) $40^\circ$ ✓
(D) $80^\circ$

Worked solution

$\text{inscribed} = \tfrac{1}{2}\,\text{arc}$ Inscribed-angle theorem.

$= \tfrac{1}{2}(80^\circ) = 40^\circ$

Answer: 40^\circ

An inscribed angle is half the central angle (and half the intercepted arc) that subtends the same arc.

Question 16 · 2 pts

An inscribed angle in a circle intercepts an arc of $80^\circ$. What is the measure of the inscribed angle?

(A) $100^\circ$
(B) $80^\circ$
(C) $160^\circ$
(D) $40^\circ$ ✓

Worked solution

$\text{inscribed}=\tfrac{1}{2}\,\text{arc}$ Inscribed-angle theorem.

$=\tfrac{1}{2}(80^\circ)=40^\circ$

Answer: 40^\circ

An inscribed angle is half the central angle on the same arc, so it equals half the intercepted arc.

Question 27 · 2 pts

Two chords intersect inside a circle. The two arcs intercepted by the pair of vertical angles measure $70^\circ$ and $30^\circ$. What is the measure of the angle formed at the intersection?

(A) $40^\circ$
(B) $100^\circ$
(C) $50^\circ$ ✓
(D) $20^\circ$

Worked solution

$\text{angle} = \tfrac{1}{2}(\text{arc}_1 + \text{arc}_2)$ Two chords meeting inside a circle

$= \tfrac{1}{2}(70^\circ + 30^\circ)$

$= \tfrac{1}{2}(100^\circ) = 50^\circ$

Answer: $50^\circ$

An angle formed by two chords intersecting inside a circle equals half the sum of the two intercepted arcs.

Question 28 · 2 pts

From an external point, two secants are drawn to a circle. They intercept a far arc of $140^\circ$ and a near arc of $50^\circ$. What is the measure of the angle at the external point?

(A) $90^\circ$
(B) $45^\circ$ ✓
(C) $95^\circ$
(D) $190^\circ$

Worked solution

$\text{angle} = \tfrac{1}{2}(\text{far arc} - \text{near arc})$ Two secants from an external point

$= \tfrac{1}{2}(140^\circ - 50^\circ)$

$= \tfrac{1}{2}(90^\circ) = 45^\circ$

Answer: $45^\circ$

An angle formed by two secants meeting at an external point equals half the positive difference of the intercepted arcs.

Question 29 · 2 pts

Chords $AB$ and $CD$ intersect at point $E$ inside a circle. $AE = 8$, $EB = 3$, and $CE = 6$. Find $ED$.

(A) $16$
(B) $4$ ✓
(C) $18$
(D) $3$

Worked solution

$AE\cdot EB = CE\cdot ED$ Intersecting chords: the products of the two segments of each chord are equal

$8\cdot 3 = 6\cdot ED$

$24 = 6\,ED \Rightarrow ED = 4$

Answer: $4$

When two chords intersect inside a circle, the product of the segments of one chord equals the product of the segments of the other.

Question 30 · 2 pts

A sector of a circle has radius $12$ and a central angle of $150^\circ$. What is the area of the sector?

(A) $60\pi$ ✓
(B) $10\pi$
(C) $120\pi$
(D) $72\pi$

Worked solution

$\text{Sector area} = \frac{\theta}{360^\circ}\cdot \pi r^2$ Sector is a fraction of the circle's area

$= \frac{150}{360}\cdot \pi(12)^2$

$= \frac{5}{12}\cdot 144\pi = 60\pi$

Answer: $60\pi$

Sector area is the fraction $\frac{\theta}{360^\circ}$ of the full circle area $\pi r^2$.

Question 18 · 2 pts

In circle $O$, points $A$, $B$, and $C$ lie on the circle, and the central angle $AOB$ measures $84°$. Point $C$ lies on the major arc $AB$. What is the measure of inscribed angle $ACB$?

(A) $84°$
(B) $138°$
(C) $21°$
(D) $42°$ ✓

Worked solution

$m\overarc{AB} = m\angle AOB = 84°$ A central angle equals its intercepted arc.

$m\angle ACB = \frac{1}{2} \, m\overarc{AB}$ An inscribed angle is half its intercepted arc. Since C is on the major arc, angle ACB intercepts minor arc AB.

$m\angle ACB = \frac{1}{2}(84°) = 42°$

Answer: $42°$

The central angle AOB equals its intercepted minor arc AB, which is 84 degrees. Since C is on the major arc, the inscribed angle ACB intercepts the minor arc AB and measures half of it, 42 degrees. The distractor 138 comes from using half of the major arc, and 84 comes from equating inscribed and central angles.

Question 28 · 4 pts · constructed response

In circle $O$, chords $\overline{AB}$ and $\overline{CD}$ intersect at point $E$ inside the circle. If $AE = 4$, $EB = 9$, and $CE = 6$, determine the length of $\overline{ED}$. Justify your answer.

Model answer: $ED = 6$. When two chords intersect inside a circle, the products of the segments of each chord are equal: $(4)(9) = (6)(ED)$, so $ED = 6$.

Worked solution

$AE \cdot EB = CE \cdot ED$ Intersecting chords theorem: the products of the two segments of each chord are equal.

$(4)(9) = (6)(ED)$

$36 = 6 \cdot ED$

$ED = 6$

Answer: $ED = 6$. When two chords intersect inside a circle, the products of the segments of each chord are equal: $(4)(9) = (6)(ED)$, so $ED = 6$.

When two chords intersect inside a circle, the product of the segments of one chord equals the product of the segments of the other. Here $4 \times 9 = 36$, so $ED = 36 \div 6 = 6$.

Question 5 · 2 pts

In a circle whose radius is $6$, a central angle measures $120°$. What is the length of the arc intercepted by this central angle, in terms of $\pi$?

(A) $2\pi$
(B) $12\pi$
(C) $36\pi$
(D) $4\pi$ ✓

Worked solution

$\text{arc length} = \frac{\theta}{360} \cdot 2\pi r$ An arc is a fraction of the full circumference.

$= \frac{120}{360} \cdot 2\pi(6)$

$= \frac{1}{3} \cdot 12\pi = 4\pi$

Answer: $4\pi$

A $120°$ central angle intercepts $\frac{120}{360} = \frac{1}{3}$ of the circle. The circumference is $2\pi(6) = 12\pi$, so the arc length is $\frac{1}{3}(12\pi) = 4\pi$. Choosing $12\pi$ is the whole circumference, and $12\pi$ is also the sector area, both common mix-ups.

Question 19 · 2 pts

In a circle, chords $\overline{AB}$ and $\overline{CD}$ intersect at point $E$ inside the circle. If arc $AC$ measures $60^\circ$ and arc $BD$ measures $100^\circ$, what is $m\angle AEC$?

(A) $20^\circ$
(B) $100^\circ$
(C) $160^\circ$
(D) $80^\circ$ ✓

Worked solution

$m\angle AEC = \frac{1}{2}\left(\text{arc } AC + \text{arc } BD\right)$ An angle formed by two chords intersecting inside a circle is half the sum of the intercepted arcs.

$m\angle AEC = \frac{1}{2}(60 + 100)$

$m\angle AEC = \frac{1}{2}(160) = 80^\circ$

Answer: $80^\circ$

When two chords intersect inside a circle, the measure of the angle formed equals half the sum of the two arcs intercepted by the angle and its vertical angle. Here that is $\frac{1}{2}(60 + 100) = 80^\circ$. Taking half the difference, $20^\circ$, applies only to angles formed outside the circle.

Question 29 · 4 pts · constructed response

In a circle whose radius is $6$ centimeters, a central angle measures $135^\circ$. Determine and state, in terms of $\pi$, the area of the sector formed by this central angle, in square centimeters.

Model answer: $13.5\pi$ square centimeters (equivalently $\frac{27\pi}{2}$).

Worked solution

$A_{\text{sector}} = \frac{135}{360} \cdot \pi r^2$ A sector's area is the fraction of the circle determined by the central angle.

$A_{\text{sector}} = \frac{135}{360} \cdot \pi (6)^2 = \frac{3}{8} \cdot 36\pi$

$A_{\text{sector}} = 13.5\pi \text{ cm}^2$

Answer: $13.5\pi$ square centimeters (equivalently $\frac{27\pi}{2}$).

A central angle of $135^\circ$ cuts off $\frac{135}{360} = \frac{3}{8}$ of the circle. The whole circle has area $\pi(6)^2 = 36\pi$, so the sector's area is $\frac{3}{8}(36\pi) = 13.5\pi$ square centimeters.

Question 10 · 2 pts

In circle $O$, points $A$, $B$, and $C$ lie on the circle, and central angle $AOC$ measures $100°$. Point $B$ lies on the major arc $AC$. What is the measure of inscribed angle $ABC$?

(A) $130°$
(B) $50°$ ✓
(C) $25°$
(D) $100°$

Worked solution

$\text{Central angle } AOC = 100° \text{, so minor arc } AC = 100°.$

$\text{Since } B \text{ is on the major arc, inscribed angle } ABC \text{ intercepts minor arc } AC.$

$m\angle ABC = \frac{1}{2} \cdot 100° = 50°$

Answer: $50°$

A central angle equals its intercepted arc, so minor arc AC measures 100 degrees. An inscribed angle measures half its intercepted arc, so angle ABC is 50 degrees. Choosing 100 confuses inscribed with central angles, and 130 comes from halving the major arc of 260 degrees.

Question 32 · 4 pts · constructed response

In circle $O$, radii $\overline{OA}$ and $\overline{OB}$ are drawn such that $m\angle AOB = 120^\circ$ and $OA = 6$ centimeters. Determine and state, to the nearest tenth of a square centimeter, the area of the sector bounded by $\overline{OA}$, $\overline{OB}$, and minor arc $AB$.

Model answer: $12\pi \approx 37.7$ square centimeters

Worked solution

$\text{Area of sector} = \frac{\theta}{360} \cdot \pi r^2$ The sector is a fraction of the whole circle determined by the central angle.

$= \frac{120}{360} \cdot \pi (6)^2$

$= \frac{1}{3} \cdot 36\pi = 12\pi$

$\approx 37.7 \text{ cm}^2$

Answer: $12\pi \approx 37.7$ square centimeters

A central angle of $120^\circ$ cuts off $\frac{120}{360} = \frac{1}{3}$ of the circle. The full circle has area $\pi(6)^2 = 36\pi$, so the sector's area is $\frac{1}{3}(36\pi) = 12\pi \approx 37.7$ square centimeters.

Question 6 · 2 pts

In a circle with radius $9$, a central angle measures $80°$. What is the length of the arc intercepted by this central angle, in terms of $\pi$?

(A) $18\pi$
(B) $4\pi$ ✓
(C) $2\pi$
(D) $8\pi$

Worked solution

$\text{arc length} = \frac{\theta}{360} \cdot 2\pi r$ The arc is the fraction of the circumference cut off by the central angle.

$= \frac{80}{360} \cdot 2\pi (9)$

$= \frac{2}{9} \cdot 18\pi = 4\pi$

Answer: $4\pi$

Arc length is the central angle's fraction of 360 degrees times the circumference. The circumference is 18 pi, and 80/360 = 2/9 of 18 pi is 4 pi. Choosing 18 pi is using the whole circumference, and 8 pi comes from doubling the correct fraction or using the diameter incorrectly.

Question 17 · 2 pts

Points $A$, $B$, and $C$ lie on a circle. Inscribed angle $ABC$ measures $36°$. What is the measure of arc $AC$, the arc intercepted by angle $ABC$?

(A) $36°$
(B) $72°$ ✓
(C) $144°$
(D) $18°$

Worked solution

$m\angle ABC = \frac{1}{2}\, m\overset{\frown}{AC}$ An inscribed angle is half its intercepted arc.

$36° = \frac{1}{2}\, m\overset{\frown}{AC}$

$m\overset{\frown}{AC} = 72°$

Answer: $72°$

The inscribed angle theorem says an inscribed angle measures half of its intercepted arc. Since the angle is $36°$, the intercepted arc $AC$ measures $2 \times 36° = 72°$. Choosing $18°$ halves the angle again instead of doubling, and $36°$ treats the arc as equal to the angle.

Question 28 · 4 pts · constructed response

In circle $O$, the radius is 6 centimeters and central angle $AOB$ measures $80^\circ$. Determine and state, to the nearest tenth of a centimeter, the length of arc $AB$ intercepted by the central angle.

Model answer: The arc length is $\frac{8\pi}{3} \approx 8.4$ centimeters.

Worked solution

$s = \frac{\theta}{360^\circ} \cdot 2\pi r$ Arc length is the fraction of the circumference cut off by the central angle.

$s = \frac{80}{360} \cdot 2\pi(6)$

$s = \frac{2}{9} \cdot 12\pi = \frac{8\pi}{3}$

$s \approx 8.4 \text{ cm}$

Answer: The arc length is $\frac{8\pi}{3} \approx 8.4$ centimeters.

An arc's length is proportional to its central angle: the $80^\circ$ angle intercepts $\frac{80}{360} = \frac{2}{9}$ of the full circumference $2\pi(6) = 12\pi$, giving $\frac{8\pi}{3} \approx 8.4$ cm.

Question 26 · 2 pts

A circle has radius $9$. What is the length of an arc intercepted by a central angle of $40^\circ$? (Leave in terms of $\pi$.)

(A) $18\pi$
(B) $2\pi$ ✓
(C) $\tfrac{\pi}{2}$
(D) $4\pi$

Worked solution

$L=\frac{\theta}{360}\cdot 2\pi r$ Arc-length formula.

$=\frac{40}{360}\cdot 2\pi(9)=\frac{1}{9}\cdot 18\pi=2\pi$

Answer: 2\pi

Arc length is the fraction θ/360 of the circumference 2πr; (40/360)(18π) = 2π.

Question 31 · 2 pts

Quadrilateral $ABCD$ is inscribed in a circle. $m\angle A = (2x+10)^\circ$ and $m\angle C = (3x)^\circ$. What is $m\angle A$?

(A) $102^\circ$
(B) $34^\circ$
(C) $30^\circ$
(D) $78^\circ$ ✓

Worked solution

$m\angle A + m\angle C = 180^\circ$ Opposite angles of a cyclic (inscribed) quadrilateral are supplementary

$(2x+10) + 3x = 180$

$5x + 10 = 180 \Rightarrow x = 34$

$m\angle A = 2(34)+10 = 78^\circ$

Answer: $78^\circ$

Opposite angles of a quadrilateral inscribed in a circle are supplementary, summing to $180^\circ$.

Question 32 · 2 pts

Two tangents are drawn to a circle from an external point, forming a $46^\circ$ angle. What is the measure of the minor arc intercepted by the two tangents?

(A) $46^\circ$
(B) $92^\circ$
(C) $134^\circ$ ✓
(D) $226^\circ$

Worked solution

$\text{angle} = \tfrac{1}{2}(\text{major arc} - \text{minor arc})$ Angle between two tangents from an external point

$\text{major} + \text{minor} = 360^\circ,\ \text{let minor} = x$

$46 = \tfrac{1}{2}\big((360 - x) - x\big) = 180 - x$

$x = 180 - 46 = 134^\circ$

Answer: $134^\circ$

The angle between two tangents from an external point equals half the difference of the intercepted arcs, and those two arcs together total $360^\circ$.

Circles & arcs: Geometry Regents Practice

More Geometry topics

More Geometry help