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Circles

Chords, tangents, inscribed angles, power of a point, arc length, sectors, inscribed/circumscribed circles

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CirclesAMC 10 Step by Step
Name:                             Date: October 28, 2026
20 problems, easiest first

What you need to know

  • Inscribed Angle Theorem. An inscribed angle (corner on the circle) is half the arc it cuts off. So an inscribed angle whose sides pass through the two ends of a diameter is a right angle. Example: an inscribed angle on a 140∘140^\circ arc is 70∘70^\circ.
  • Tangents. A tangent line is perpendicular to the radius at the point where it touches. Two Tangent Theorem: the two tangent segments from a point outside the circle are equal. Example: from a point 1010 away from the center of a circle with radius 66, both tangents are 100−36=8\sqrt{100-36} = 8 long.
  • Power of a Point. If chords ABAB and CDCD cross at PP, then PA⋅PB=PC⋅PDPA\cdot PB = PC\cdot PD. From a point PP outside the circle, with a tangent PTPT and a line (a secant) meeting the circle at AA and then BB: PT2=PA⋅PBPT^2 = PA\cdot PB. Example: chords with PA=4PA = 4, PB=6PB = 6, PC=3PC = 3 give PD=8PD = 8. A secant with PA=4PA = 4, PB=9PB = 9 gives PT=6PT = 6.
  • Arcs and sectors. A central angle of θ\theta degrees cuts off θ360∘\frac{\theta}{360^\circ} of the circle: the arc length is θ360∘⋅2πr\frac{\theta}{360^\circ}\cdot 2\pi r, and the sector (pizza slice) has area θ360∘⋅πr2\frac{\theta}{360^\circ}\cdot \pi r^2. Example: with r=6r = 6 and θ=60∘\theta = 60^\circ, the arc is 2π2\pi long and the sector's area is 6π6\pi.

How AMC 10 tests it

  • Shaded regions (problems 3–12): circles in squares, or semicircles on the sides of a triangle. Subtract sectors and triangles.
  • Tangent lines from a point, or one line tangent to two circles. Radii to the points of tangency give a right triangle or a trapezoid with two right angles.
  • Problems 17–23: Power of a Point for a length, or a rope wrapping around a polygon, giving a sum of sectors.

Standard approaches

  1. Draw the radius to every point of tangency and every chord end. Look for right angles, and for isosceles triangles (radii are equal).
  2. Turn inscribed angles into arcs, add the arcs, and turn back.
  3. When two chords, or a tangent and a secant, pass through one point, write Power of a Point.

Worked example

Point PP lies outside a circle with center OO and radius 66. The two tangent segments from PP touch the circle at AA and BB, and ∠APB=60∘\angle APB = 60^\circ. What is the area of the region bounded by PA‾\overline{PA}, PB‾\overline{PB}, and the minor arc ABAB?

(A) 183−6π18\sqrt3 - 6\pi (B) 363−6π36\sqrt3 - 6\pi (C) 363−12π36\sqrt3 - 12\pi (D) 36−12π36 - 12\pi (E) 723−12π72\sqrt3 - 12\pi

Solution. Step 1. Each tangent is perpendicular to its radius, so ∠OAP=∠OBP=90∘\angle OAP = \angle OBP = 90^\circ.

Step 2. Right triangles OAPOAP and OBPOBP share the hypotenuse OPOP and have OA=OBOA = OB, so they are congruent, and ∠APO=12⋅60∘=30∘\angle APO = \frac12\cdot60^\circ = 30^\circ. Triangle OAPOAP is a 3030-6060-9090 triangle with short leg OA=6OA = 6, so PA=63PA = 6\sqrt3 and ∠AOP=60∘\angle AOP = 60^\circ.

Step 3. The kite OAPBOAPB is two copies of triangle OAPOAP, so its area is 2⋅12⋅6⋅63=3632\cdot\frac12\cdot 6\cdot 6\sqrt3 = 36\sqrt3.

Step 4. The central angle is ∠AOB=120∘\angle AOB = 120^\circ, so sector OABOAB has area 120360⋅36π=12π\frac{120}{360}\cdot 36\pi = 12\pi.

Step 5. The region is the kite minus the sector: 363−12π36\sqrt3 - 12\pi.

The answer is (C) 363−12π\boxed{\textbf{(C)}\ 36\sqrt3 - 12\pi}.

Pitfalls

  • Putting the diameter into πr2\pi r^2 instead of the radius. The area comes out 44 times too big, and the answer choices often differ by that factor of 44.
  • Finding a sector's area with the inscribed angle instead of the central angle (which is twice as big).
  • In Power of a Point, using PA⋅ABPA\cdot AB instead of PA⋅PBPA\cdot PB. Both lengths start at PP. Example: with PA=4PA = 4 and AB=5AB = 5, the product is 4⋅9=364\cdot 9 = 36, not 2020.

Traps that recur

  • Assuming the inner corners lie on a circle of radius 3 (the table radius minus the mat width), rather than at distance x from the center. (2008 AMC 10A #25)
  • Assuming the inner circle is the incircle of the triangle of centres, whose inradius is 1, or reading the picture's apparent symmetry as meaning the inner centre is equidistant from the three outer centres. (2025 AMC 10A #22)
  • Losing a minus sign when writing the diameter, giving (sqrt 2 - 1)(sqrt 3 + 1) and the sum 5 in (B). (2025 AMC 10B #20)
  • Finding only the small pipe wedged between the two and answering 1/9, forgetting that a pipe on the far side of the small one is also in contact with both. (2024 AMC 10B #21)
  1. 1.2008 AMC 10B #10

    Points AA and BB are on a circle of radius 55 and AB=6AB=6 . Point CC is the midpoint of the minor arc ABAB . What is the length of the line segment ACAC ?

    1. A)10\sqrt{10}
    2. B)72\frac{7}{2}
    3. C)14\sqrt{14}
    4. D)15\sqrt{15}
    5. E)44
  2. 2.2007 AMC 10B #4

    The point OO is the center of the circle circumscribed about △ABC,\triangle ABC, with ∠BOC=120∘\angle BOC=120^\circ and ∠AOB=140∘,\angle AOB=140^\circ, as shown. What is the degree measure of ∠ABC?\angle ABC?

    1. A)3535
    2. B)4040
    3. C)4545
    4. D)5050
    5. E)6060
  3. 3.2007 AMC 10A #14

    A triangle with side lengths in the ratio 3 :4 :53 : 4 : 5 is inscribed in a circle with radius 3. What is the area of the triangle?

    1. A)8.648.64
    2. B)1212
    3. C)5π5\pi
    4. D)17.2817.28
    5. E)1818
  4. 4.2006 AMC 10B #8

    A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?

    figure

    1. A)20π20\pi
    2. B)25π25\pi
    3. C)30π30\pi
    4. D)40π40\pi
    5. E)50π50\pi
  5. 5.2006 AMC 10A #12

    Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.

    figure

    Which of these arrangements give the dog the greater area to roam, and by how many square feet?

    1. A)I, by 8πI, \textrm{ by } 8\pi
    2. B)I, by 6πI, \textrm{ by } 6\pi
    3. C)II, by 4πII, \textrm{ by } 4\pi
    4. D)II, by 8πII, \textrm{ by } 8\pi
    5. E)II, by 10πII, \textrm{ by } 10\pi
  6. 6.2005 AMC 10B #7

    A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?

    1. A)π16\frac{\pi}{16}
    2. B)π8\frac{\pi}{8}
    3. C)3π16\frac{3\pi}{16}
    4. D)π4\frac{\pi}{4}
    5. E)π2\frac{\pi}{2}
  7. 7.2005 AMC 10B #8

    An 88 -foot by 1010 -foot bathroom floor is tiled with square tiles of size 11 foot by 11 foot. Each tile has a pattern consisting of four white quarter circles of radius 1/21/2 foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?

    figure

    1. A)80−20π80-20\pi
    2. B)60−10π60-10\pi
    3. C)80−10π80-10\pi
    4. D)60+10π60+10\pi
    5. E)80+10π80+10\pi
  8. 8.2004 AMC 10B #9

    A square has sides of length 1010 , and a circle centered at one of its vertices has radius 1010 . What is the area of the union of the regions enclosed by the square and the circle?

    1. A)200+25π200+25\pi
    2. B)100+75π100+75\pi
    3. C)75+100π75+100\pi
    4. D)100+100π100+100\pi
    5. E)100+125π100+125\pi
  9. 9.2004 AMC 10B #12

    An annulus is the region between two concentric circles. The concentric circles in the figure have radii bb and cc , with b>cb>c . Let OXOX be a radius of the larger circle, let XZXZ be tangent to the smaller circle at ZZ , and let OYOY be the radius of the larger circle that contains ZZ . Let a=XZa=XZ , d=YZd=YZ , and e=XYe=XY . What is the area of the annulus?

    figure

    1. A)πa2\pi a^2
    2. B)πb2\pi b^2
    3. C)πc2\pi c^2
    4. D)πd2\pi d^2
    5. E)πe2\pi e^2
  10. 10.2002 AMC 10A #7

    A 45∘45^\circ arc of circle A is equal in length to a 30∘30^\circ arc of circle B. What is the ratio of circle A's area and circle B's area?

    1. A)4/94/9
    2. B)2/32/3
    3. C)5/65/6
    4. D)3/23/2
    5. E)9/49/4
  11. 11.2025 AMC 10B #12

    The figure below shows an equilateral triangle, a rhombus with a 60∘60^\circ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let T,R,T, R, and H,H, respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon. Which of the following is true?

    figure

    1. A)T=H=RT=H=R
    2. B)H<R=TH<R=T
    3. C)H=R<TH=R<T
    4. D)H<R<TH<R<T
    5. E)H<T<RH<T<R
  12. 12.2025 AMC 10A #10

    A semicircle has diameter AB‾\overline{AB} and chord CD‾\overline{CD} of length 1616 parallel to AB‾\overline{AB} . A smaller semicircle with diameter on AB‾\overline{AB} and tangent to CD‾\overline{CD} is cut from the larger semicircle, as shown below.

    figure

    What is the area of the resulting figure, shown shaded?

    1. A)16π16\pi
    2. B)24π24\pi
    3. C)32π32\pi
    4. D)48π48\pi
    5. E)64π64\pi
  13. 13.2024 AMC 10A #14

    One side of an equilateral triangle of height 2424 lies on line ℓ\ell . A circle of radius 1212 is tangent to line ℓ\ell and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line ℓ\ell can be written as ab−cπa \sqrt{b} - c \pi , where aa , bb , and cc are positive integers and bb is not divisible by the square of any prime. What is a+b+ca + b + c ?

    1. A)7272
    2. B)7373
    3. C)7474
    4. D)7575
    5. E)7676
  14. 14.2023 AMC 10A #13

    Abdul and Chiang are standing 4848 feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures 60∘60^\circ . What is the square of the distance (in feet) between Abdul and Bharat?

    1. A)17281728
    2. B)26012601
    3. C)30723072
    4. D)46084608
    5. E)69126912
  15. 15.2022 AMC 10A #15

    Quadrilateral ABCDABCD with side lengths AB=7,BC=24,CD=20,DA=15AB=7, BC=24, CD=20, DA=15 is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form aπ−bc,\frac{a\pi-b}{c}, where a,b,a,b, and cc are positive integers such that aa and cc have no common prime factor. What is a+b+c?a+b+c?

    1. A)260260
    2. B)855855
    3. C)12351235
    4. D)15651565
    5. E)19971997
  16. 16.2021 AMC Fall 10A #15

    Isosceles triangle ABCABC has AB=AC=36AB = AC = 3\sqrt6 , and a circle with radius 525\sqrt2 is tangent to line ABAB at BB and to line ACAC at CC . What is the area of the circle that passes through vertices AA , BB , and C?C?

    1. A)24π24\pi
    2. B)25π25\pi
    3. C)26π26\pi
    4. D)27π27\pi
    5. E)28π28\pi
  17. 17.2021 AMC Fall 10B #11

    A regular hexagon of side length 11{ } is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these 66 reflected arcs?

    1. A)532−π\frac{5\sqrt{3}}{2} - \pi
    2. B)33−π3\sqrt{3}-\pi
    3. C)43−3π24\sqrt{3}-\frac{3\pi}{2}
    4. D)π−32\pi - \frac{\sqrt{3}}{2}
    5. E)π+32\frac{\pi + \sqrt{3}}{2}
  18. 18.2021 AMC 10B #14

    Three equally spaced parallel lines intersect a circle, creating three chords of lengths 38,38,38, 38, and 3434 . What is the distance between two adjacent parallel lines?

    1. A)5125\frac{1}{2}
    2. B)66
    3. C)6126\frac{1}{2}
    4. D)77
    5. E)7127\frac{1}{2}
  19. 19.2020 AMC 10B #14

    As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?

    figure

    1. A)63−3π6\sqrt{3}-3\pi
    2. B)932−2π\frac{9\sqrt{3}}{2} - 2\pi
    3. C)332−π3\frac{3\sqrt{3}}{2} - \frac{\pi}{3}
    4. D)33−π3\sqrt{3} - \pi
    5. E)932−π\frac{9\sqrt{3}}{2} - \pi
  20. 20.2019 AMC 10A #16

    The figure below shows 1313 circles of radius 11 within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius 1?1 ?

    figure

    1. A)4π34 \pi \sqrt{3}
    2. B)7π7 \pi
    3. C)π(33+2)\pi\left(3\sqrt{3} +2\right)
    4. D)10π(3−1)10 \pi \left(\sqrt{3} - 1\right)
    5. E)π(3+6)\pi\left(\sqrt{3} + 6\right)
Circles — answersAMC 10 Step by Step
Name:                             Date: October 28, 2026
Keep this page separate from the problems.

Answers

#AnswerThe idea it turns on
1ADraw a line from the center through the chord's midpoint. The chord is 4 from the center and C is 1 beyond it, so AC has legs 3 and 1.
2DThe three center angles add to 360, so angle AOC is 100. The angle at B looks at the same arc, so it is half of that.
3AA 3-4-5 triangle has a right angle, so its longest side is a diameter of the circle, which is 6. Scale the 3-4-5 triangle up by 6/5.
4BA line from the center to a top corner is a radius: the long side of a right triangle with legs equal to the side and half the side.
5CBoth drawings give the same half-circle of radius 8. In II the rope reaches past the nearby corner with 4 feet left over, adding a quarter-circle of radius 4.
6BA circle in a square has diameter equal to the side; a square in a circle has diagonal equal to the diameter. Start with side 2.
7AOn one tile, the four white quarter circles fit together into one whole circle of radius 1/2. So each tile has 1 minus pi/4 shaded, and there are 80 tiles.
8BThe circle is centered at a corner of the square, so exactly a quarter of the circle overlaps the square. Add the square and the other three quarters.
9AA radius meets a tangent line at a right angle, so triangle OZX is a right triangle. The Pythagorean theorem gives b squared minus c squared equals a squared.
10ACircle A needs a bigger angle for the same arc, so it is smaller. The radii are in ratio 2 to 3, and areas square that: 4 to 9.
11CCut the hexagon into six equilateral triangles and the rhombus into two: each disk is the circle inside one triangle, so H = R.
12CThe small radius equals the height of the chord above AB. A right triangle with legs 8 and that height gives big radius squared minus small radius squared = 64.
13DThe circle sits in the 120 degree corner outside the triangle. The region is a kite minus a 60 degree slice of the circle.
14CAll the spots where Bharat sees Abdul and Chiang at a 60 degree angle lie on one circle through them. Abdul-to-Bharat is longest when it is a diameter.
15D7, 24, 25 and 15, 20, 25 are both right-triangle side sets with long side 25. So the diagonal AC is 25 and is a diameter of the circle.
16CTangent radii make right angles at B and C, so right triangles ABO and ACO share side AO. The circle through A, B, C has AO as diameter.
17BEach reflected arc bends into the hexagon, so the region is the hexagon minus six thin slivers; each sliver is a 60-degree pie slice minus an equilateral triangle.
18BEqual chords are equally far from the center, so it sits halfway between the 38-chords. Then 19² + (d/2)² = 17² + (3d/2)².
19DNeighboring semicircles cross halfway between the center and each corner. The shaded part is six small triangles of side 1, each missing a curved bite.
20AJoin the centers of touching circles; each join is 2 long. The top center is 2 times root 3 above the middle, so the big radius is that plus 1.

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.