Circles

Study Sheet

Circles

Parts, circumference, area, angles, tangents, chords, and equations

The Parts of a Circle

Concept
Naming the pieces

A circle is the set of all points that are the same distance from one point, the center. We name a circle by its center: a circle with center OO is “circle OO,” written ⊙O\odot O.

  • Radius: a segment from the center to a point on the circle.
  • Diameter: a chord that passes through the center. It is the longest chord, and d=2rd = 2r.
  • Chord: a segment whose two endpoints are on the circle.
  • Secant: a line that cuts the circle at two points.
  • Tangent: a line that touches the circle at exactly one point.

Circle theorems come in three families: angles (a central angle equals its arc, an inscribed angle is half its arc, an angle in a semicircle is right), lengths (equal chords are equidistant, a tangent is perpendicular to the radius, power of a point), and measurements (arc length and sector area are fractions of the circumference and the area).

Example
A circle with its interior parts labeled
OABCDE
OC is a radius, AB a diameter through O, DE a chord that misses O.

Here OC‾\overline{OC} is a radius, AB‾\overline{AB} is a diameter (it goes through OO), and DE‾\overline{DE} is a chord (its endpoints are on the circle, but it misses the center).

Side note
Where it goes wrong

Using the inscribed angle as equal to its arc instead of half. Central angle == arc; inscribed angle == half the arc.

Example
Secant line vs. tangent line
Psecanttangent
The secant crosses at two points; the tangent touches only at P.

A secant slices through the circle at two points. A tangent just grazes it at one point, PP (the point of tangency).

Concept
Arcs: pieces of the circle itself

An arc is a curved piece of the circle between two points.

  • Minor arc: the shorter way around (less than 180∘180^\circ). Named with two letters, like AB⌢\overset{\frown}{AB}.
  • Major arc: the longer way around (more than 180∘180^\circ). Named with three letters to show which way you travel.
  • Semicircle: exactly half the circle (180∘180^\circ); its endpoints are the ends of a diameter.
Example
Minor arc and major arc
80°ABC
Solid: minor arc AB (80∘^\circ). Dashed: major arc ACB (280∘^\circ).

Points AA and BB split the circle into two arcs. The short (solid) path is the minor arc AB⌢\overset{\frown}{AB}; the long (dashed) way around is the major arc, named ACB⌢\overset{\frown}{ACB} with a third point CC on it so the reader knows which way to travel.

Tip

Radius vs. diameter. The diameter is always twice the radius: d=2rd = 2r, so r=12dr = \tfrac{1}{2}d. Mixing these up is the most common circle mistake — always ask “do I have the radius or the diameter?”

Try it
Quick check

A chord of a circle has length 1010. Could the radius be 44? Could it be 55?

Show the answer

The longest chord is a diameter, 2r2r. With r=4r = 4 every chord is at most 88, so a chord of 1010 is impossible. With r=5r = 5 the diameter is exactly 1010, so it is possible — the chord must be a diameter.

Circumference and Arc Length

Concept
Circumference: the distance around
510π ≈ 31.4
Unrolled, the circle of radius 5 is a little over three diameters long (the faint ticks mark 10, 20, 30).

The circumference CC is the perimeter of a circle:

C=2πr=πd.C = 2\pi r = \pi d .

The number π\pi (“pi”) is about 3.143.14. We leave answers in terms of π\pi unless a decimal is asked for; then we use π≈3.14\pi \approx 3.14.

Every circle is a scaled copy of every other, so the ratio circumference : diameter is the same for all of them; that constant ratio is what π\pi means. Hence C=πd=2πrC = \pi d = 2\pi r. Arc length is the same idea applied to a slice: an arc with central angle n∘n^\circ is n360\tfrac{n}{360} of the full turn, so it is that fraction of the circumference.

Example
Circumference from the radius

A circle has radius r=5r = 5 cm.

C=2πr=2π(5)=10π cm≈10(3.14)=31.4 cm.C = 2\pi r = 2\pi(5) = 10\pi \text{ cm} \approx 10(3.14) = 31.4 \text{ cm}.
Concept
Arc length: part of the way around

An arc is a fraction of the whole circle. If its central angle is n∘n^\circ, then

arc length=n360⋅2πr.\text{arc length} = \frac{n}{360}\cdot 2\pi r .

The fraction n360\dfrac{n}{360} is “how much of the full 360∘360^\circ” the arc covers.

Example
Arc length
84π
A 90∘^\circ arc is a quarter of the circumference 16π\pi.

Find the length of an arc with central angle 90∘90^\circ in a circle of radius 88.

arc length=90360⋅2π(8)=14⋅16π=4π≈12.56.\text{arc length} = \frac{90}{360}\cdot 2\pi(8) = \frac{1}{4}\cdot 16\pi = 4\pi \approx 12.56 .
Side note
Where it goes wrong

Using the diameter where the radius belongs: 90360⋅2π(16)=8π\tfrac{90}{360} \cdot 2\pi(16) = 8\pi when the diameter is 1616 counts the circle twice over. The formula n360⋅2πr\tfrac{n}{360}\cdot 2\pi r wants the radius; with a diameter use n360⋅πd\tfrac{n}{360}\cdot\pi d.

Tip

Watch the units. Circumference and arc length are lengths, so they use ordinary units (cm, m, in). Later, area will use square units. If you see “cm2^2” on a length answer, something went wrong.

Try it
Quick check

Find the length of an arc with central angle 135∘135^\circ in a circle of radius 1212. Give an exact answer and a decimal.

Show the answer

135360⋅2π(12)=38⋅24π=9π≈28.27\dfrac{135}{360} \cdot 2\pi(12) = \dfrac38 \cdot 24\pi = \mathbf{9\pi} \approx 28.27.

Central Angles and Arc Measure

Concept
A central angle equals its arc

A central angle has its vertex at the center of the circle. The measure of a minor arc equals the measure of its central angle. A full trip around the circle is 360∘360^\circ, so all the central angles around the center add to 360∘360^\circ.

This is really a definition: we measure an arc by the central angle that cuts it off, so a full circle is 360∘360^\circ of arc. Arc measure says how much of the turn the arc covers, not how long it is — a 60∘60^\circ arc on a small circle and on a large one have the same measure but different lengths.

Example
Central angle == arc
60°ABO300°
Minor arc AB = 60∘^\circ (solid); the major arc is 300∘^\circ (dashed).

The central angle ∠AOB\angle AOB measures 60∘60^\circ, so the intercepted arc AB⌢\overset{\frown}{AB} also measures 60∘60^\circ. The major arc is 360∘−60∘=300∘360^\circ - 60^\circ = 300^\circ.

Side note
Where it goes wrong

Confusing arc measure with arc length. "Arc AB=60∘AB = 60^\circ" is a measure in degrees; its length in a circle of radius 33 is 60360⋅6π=π\tfrac{60}{360}\cdot 6\pi = \pi units. Giving "6060 cm" as a length mixes the two.

Tip

Everything adds to 360∘360^\circ. If a circle is cut into central angles, their measures total 360∘360^\circ. To find a missing central angle, subtract the known ones from 360∘360^\circ.

Try it
Quick check

Four central angles fill a circle and measure xx, 2x2x, 3x3x and 4x4x degrees. Find the four arcs.

Show the answer

10x=36010x = 360, so x=36x = 36. The arcs are 36∘,72∘,108∘,144∘\mathbf{36^\circ, 72^\circ, 108^\circ, 144^\circ}.

Inscribed Angles

Concept
The Inscribed Angle Theorem

An inscribed angle has its vertex on the circle, and its two sides are chords. The

inscribed angle=12 (intercepted arc).\text{inscribed angle} = \tfrac{1}{2}\,(\text{intercepted arc}).

So the intercepted arc is twice the inscribed angle. (Notice a central angle equals its arc, but an inscribed angle is only half of it.)

Why half: when one side of the inscribed angle passes through the center, the radius to the other endpoint makes an isosceles triangle, and the central angle is an exterior angle of that triangle — so it equals the two equal base angles together, twice the inscribed angle. Every other position is a sum or difference of two such cases (the proof is in Going Deeper). A consequence worth knowing: all inscribed angles on the same arc are equal, wherever their vertex sits on the rest of the circle.

Example
Central angle vs. inscribed angle on the same arc
90°45°ABCO
Both angles open onto arc AB: the central angle is 90∘^\circ, the inscribed one 45∘^\circ.

Both angles below open onto arc AB⌢\overset{\frown}{AB}. The central angle ∠AOB=90∘\angle AOB = 90^\circ equals the arc. The inscribed angle ∠ACB\angle ACB is half of that arc: 12(90∘)=45∘\tfrac{1}{2}(90^\circ) = 45^\circ.

Example
Angle inscribed in a semicircle is 90∘90^\circ
ABCO
AB is a diameter, so the angle at C is right, wherever C is.

If a chord is a diameter, the arc it cuts off is a semicircle (180∘180^\circ). Any inscribed angle on that arc is 12(180∘)=90∘\tfrac{1}{2}(180^\circ) = 90^\circ. So ∠ACB\angle ACB below is a right angle.

Side note
Where it goes wrong

Taking the inscribed angle equal to its arc. An inscribed angle on a 100∘100^\circ arc is 50∘50^\circ; only a central angle equals its arc.

Tip

Two big ideas. (1) Inscribed angle == half its arc. (2) An angle inscribed in a semicircle (sitting on a diameter) is always 90∘90^\circ. This second fact turns diameters into right triangles!

Reminder — Inscribed angle theorem:  inscribed angle=12 central angle on the same arc\text{inscribed angle}=\tfrac12\,\text{central angle on the same arc}
Try it
Quick check

An inscribed angle measures 38∘38^\circ. Find its intercepted arc and the central angle on the same arc.

Show the answer

The arc is twice the inscribed angle: 76∘\mathbf{76^\circ}. The central angle equals the arc: 76∘\mathbf{76^\circ}.

Tangent Lines

Concept
Tangent ⊥\perp radius

A tangent touches a circle at exactly one point. At that point of tangency, the tangent line is perpendicular to the radius. That right angle is the key to almost every tangent problem, because it makes a right triangle you can use with the Pythagorean Theorem.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2

Why the angle is right: of all the points on a line, the one closest to OO is the foot of the perpendicular from OO. Every point of the tangent except PP lies outside the circle, farther than rr from OO, so PP is the closest point of the line to OO — the foot of the perpendicular. If the radius met the line at any other angle, the foot would be closer than rr, inside the circle, and the line would cut the circle twice.

Example
The tangent meets the radius at 90∘90^\circ
OPt
The radius to the point of tangency is perpendicular to t.

Radius OP‾\overline{OP} meets tangent line tt at the point of tangency PP, forming a right angle.

Example
Two tangents from one outside point are equal
PT₁T₂O
Right triangles OT1_1P and OT2_2P share OP and have equal radii: congruent by HL.

From an outside point PP, the two tangent segments to the circle are congruent: PT1=PT2PT_1 = PT_2.

Example
Using the right triangle
51213OTP
OP is the hypotenuse: 52^2 + 122^2 = 132^2.

A tangent from point PP touches ⊙O\odot O at TT. If OP=13OP = 13 and radius OT=5OT = 5, find the tangent length PTPT. Because ∠OTP=90∘\angle OTP = 90^\circ:

PT=OP2−OT2=132−52=169−25=144=12.PT = \sqrt{OP^2 - OT^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 .
Side note
Where it goes wrong

Treating OPOP as a leg. The right angle is at TT, so the side opposite it, OPOP, is the hypotenuse: PT=132−52=12PT = \sqrt{13^2 - 5^2} = 12, not 132+52≈13.9\sqrt{13^2 + 5^2} \approx 13.9.

Tip

Draw the radius. Whenever a tangent appears, draw the radius to the point of tangency. You instantly get a 90∘90^\circ angle — and a right triangle where OPOP (center to outside point) is the hypotenuse.

Try it
Quick check

From an outside point PP, two tangents touch a circle at T1T_1 and T2T_2, with PT1=3x+2PT_1 = 3x + 2 and PT2=5x−8PT_2 = 5x - 8. Find PT1PT_1.

Show the answer

Tangent segments from one point are equal: 3x+2=5x−83x + 2 = 5x - 8, so x=5x = 5 and PT1=17PT_1 = \mathbf{17}.

Chord Relationships

Concept
Two chord facts
  • Equal chords: In the same circle, chords that are the same distance from the center are congruent (equal in length).
  • Perpendicular from the center: A radius (or diameter) that is perpendicular to a chord bisects it — it cuts the chord into two equal halves.

Both facts come from one pair of triangles. Draw radii to the ends of a chord AB‾\overline{AB} and the perpendicular OM‾\overline{OM}: right triangles OMAOMA and OMBOMB share OM‾\overline{OM} and have equal hypotenuses (radii), so they are congruent by HL, and AM=MBAM = MB. The same right triangle gives AM2=r2−OM2AM^2 = r^2 - OM^2: the half-chord is fixed by the distance from the center, so chords at equal distances are equal, and nearer chords are longer.

Example
The perpendicular from the center bisects the chord
354OABM
Radius 5, distance 3, half-chord 4: a 3-4-5 triangle.

OM‾⊥AB‾\overline{OM}\perp\overline{AB}, so MM is the midpoint of chord AB‾\overline{AB} and AM=MBAM = MB. This also makes a right triangle OMAOMA: if the radius OA=5OA = 5 and OM=3OM = 3, then

AM=OA2−OM2=25−9=16=4,so AB=2(4)=8.AM = \sqrt{OA^2 - OM^2} = \sqrt{25 - 9} = \sqrt{16} = 4, \quad\text{so } AB = 2(4) = 8 .
Side note
Where it goes wrong

Putting the whole chord into the right triangle: 25−9=4\sqrt{25 - 9} = 4 is half the chord, so AB=8AB = 8. Writing AB=4AB = 4, or using ABAB as a leg, is off by a factor of 22.

Tip

Half the chord, please. When you drop a perpendicular from the center to a chord, the right triangle uses half the chord as a leg and the radius as the hypotenuse. Don't plug in the whole chord by accident.

Try it
Quick check

A chord of length 2424 lies in a circle of radius 1313. How far is it from the center?

Show the answer

Half the chord is 1212, the radius 1313 is the hypotenuse, so the distance is 132−122=25=5\sqrt{13^2 - 12^2} = \sqrt{25} = \mathbf{5}.

Area of a Circle and Area of a Sector

Concept
Area of a circle
about πrr
Cut into wedges and laid alternately: nearly a parallelogram of base π\pir and height r.

The area inside a circle is

A=πr2.A = \pi r^2 .

Remember it is rr squared, then times π\pi. Area is measured in square units.

Why πr2\pi r^2: cut the circle into many thin wedges and lay them head-to-tail, alternately pointing up and down. Half the curved edges end up along the bottom and half along the top, so the base is half the circumference, πr\pi r; the height is the radius. The more wedges, the closer the shape is to a parallelogram, whose area is πr⋅r=πr2\pi r \cdot r = \pi r^2. A sector is n360\tfrac{n}{360} of the turn, so it gets that fraction of the area.

Example
Area from the radius

A circle has radius r=5r = 5.

A=πr2=π(5)2=25π≈25(3.14)=78.5 square units.A = \pi r^2 = \pi (5)^2 = 25\pi \approx 25(3.14) = 78.5 \text{ square units}.
Concept
Area of a sector
60°6
A 60∘^\circ sector is one sixth of the disc: 36π÷\pi \div 6 = 6π\pi.

A sector is a “pizza slice” bounded by two radii and an arc. If its central angle is n∘n^\circ,

sector area=n360⋅πr2.\text{sector area} = \frac{n}{360}\cdot \pi r^2 .

It is the same fraction n360\dfrac{n}{360} of the whole circle that we used for arc length.

Side note
Where it goes wrong

Squaring after multiplying by π\pi, or forgetting to square: π⋅6=6π\pi \cdot 6 = 6\pi is not the area of the whole circle; π⋅62=36π\pi \cdot 6^2 = 36\pi is, and the sector is 16\tfrac16 of that.

Example
Area of a sector

Find the area of a 60∘60^\circ sector in a circle of radius 66.

sector area=60360⋅π(6)2=16⋅36π=6π≈18.84.\text{sector area} = \frac{60}{360}\cdot \pi (6)^2 = \frac{1}{6}\cdot 36\pi = 6\pi \approx 18.84 .
Tip

Same fraction, two uses. The fraction n360\dfrac{n}{360} scales the whole circumference to get arc length, and scales the whole area to get sector area. Learn it once, use it twice!

Try it
Quick check

A sector of a circle of radius 66 has area 15π15\pi. Find its central angle.

Show the answer

The whole disc is 36π36\pi, so the sector is 15π36π=512\tfrac{15\pi}{36\pi} = \tfrac{5}{12} of it. The angle is 512⋅360∘=150∘\tfrac{5}{12} \cdot 360^\circ = \mathbf{150^\circ}.

The Equation of a Circle

Concept
Center–radius form

A circle with center (h,k)(h,k) and radius rr has the equation

(x−h)2+(y−k)2=r2.(x-h)^2 + (y-k)^2 = r^2 .

Read it carefully: the numbers subtracted from xx and yy give the center, and the right side is r2r^2 (so take the square root to get the radius).

The equation is the distance formula with the distance fixed. A point (x,y)(x, y) is on the circle exactly when its distance from the center (h,k)(h, k) is rr: (x−h)2+(y−k)2=r\sqrt{(x - h)^2 + (y - k)^2} = r. Squaring both sides (both are non-negative, so nothing is gained or lost) gives (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.

Example
Writing an equation
xy−1123456−11234562
Center (3, 3), radius 2.

A circle has center (3,3)(3,3) and radius 22. Its equation is

(x−3)2+(y−3)2=22=4.(x-3)^2 + (y-3)^2 = 2^2 = 4 .
Example
Reading center and radius from an equation
xy−3−2−1123456−8−7−6−5−4−3−2−114
(x −- 2)2^2 + (y + 3)2^2 = 16: center (2, −-3), radius 4.

Given (x−2)2+(y+3)2=16(x-2)^2 + (y+3)^2 = 16. Rewrite y+3y+3 as y−(−3)y-(-3), so the center is (2,−3)(2,-3). Since r2=16r^2 = 16, the radius is r=16=4r = \sqrt{16} = 4.

Side note
Where it goes wrong

Reading the signs as written, or forgetting the square root. (x−2)2+(y+3)2=16(x - 2)^2 + (y + 3)^2 = 16 is not centered at (−2,3)(-2, 3) and does not have radius 1616: the center is (2,−3)(2, -3) and the radius is 44.

Tip

Mind the signs. The center uses the opposite sign of what you see: (x−2)(x-2) means h=+2h = +2, and (y+3)=(y−(−3))(y+3) = (y-(-3)) means k=−3k = -3. And the right side is r2r^2, not rr — always take the square root.

Try it
Quick check

Is the point (6,−1)(6, -1) on the circle (x−2)2+(y+4)2=25(x - 2)^2 + (y + 4)^2 = 25?

Show the answer

Substitute: (6−2)2+(−1+4)2=16+9=25(6 - 2)^2 + (-1 + 4)^2 = 16 + 9 = 25. Yes — it is exactly 55 from the center (2,−4)(2, -4).

Going Deeper: Advanced Circle Ideas

Concept
Proving the Inscribed Angle Theorem
xx2xCABO
OB = OC, so triangle OBC is isosceles; the exterior angle at O is x + x.

Why is an inscribed angle exactly half its arc? Look at the easy case where one side of the angle is a diameter. Draw radius OB‾\overline{OB}. Since OB=OC=rOB = OC = r, triangle OBCOBC is isosceles, so its base angles are equal: ∠OCB=∠OBC\angle OCB = \angle OBC. Now ∠AOB\angle AOB is an exterior angle of that triangle, so it equals the sum of the two remote interior angles:

∠AOB=∠OCB+∠OBC=2 ∠OCB=2 ∠ACB.\angle AOB = \angle OCB + \angle OBC = 2\,\angle OCB = 2\,\angle ACB .

The central angle ∠AOB\angle AOB equals arc AB⌢\overset{\frown}{AB}, so the inscribed angle ∠ACB\angle ACB is half that arc. (Any inscribed angle can be split into cases like this by drawing the diameter through CC.)

Concept
Why the perpendicular from the center bisects a chord
OABM
Hypotenuse OA = OB, shared leg OM: congruent by HL.

Section on chords stated it; here is why. Drop OM‾⊥AB‾\overline{OM}\perp\overline{AB} and draw radii OA‾\overline{OA} and OB‾\overline{OB}. Then right triangles OMAOMA and OMBOMB share leg OM‾\overline{OM} and have equal hypotenuses OA=OB=rOA = OB = r. By Hypotenuse–Leg, the triangles are congruent, so AM=MBAM = MB. The perpendicular from the center must split the chord in half.

Concept
The tangent–chord angle
60°120°PBt
Chord PB cuts off a 120∘^\circ arc; the tangent-chord angle is 60∘^\circ.

When a tangent and a chord meet at the point of tangency, the angle they form is half the intercepted arc — the same “half the arc” rule as an inscribed angle:

tangent–chord angle=12 (intercepted arc).\text{tangent--chord angle} = \tfrac{1}{2}\,(\text{intercepted arc}).

In the picture, chord PB‾\overline{PB} and tangent tt meet at PP; the marked angle is half of arc PB⌢\overset{\frown}{PB}.

Concept
Cyclic quadrilaterals: opposite angles are supplementary

A cyclic quadrilateral has all four vertices on a circle. Each angle is inscribed, so it equals half of the arc across from it. The two arcs opposite a pair of angles together make the whole circle (360∘360^\circ), so each pair of opposite angles adds to half of that:

∠A+∠C=180∘,∠B+∠D=180∘.\angle A + \angle C = 180^\circ, \qquad \angle B + \angle D = 180^\circ .
Concept
The Power of a Point

If two lines through a point PP each meet the circle, the products of the two distances along each line are equal. There are three versions of this one idea:

  • Two chords crossing inside at PP:  PA⋅PB=PC⋅PD\ PA\cdot PB = PC\cdot PD.
  • Two secants from an outside point PP:  PA⋅PB=PC⋅PD\ PA\cdot PB = PC\cdot PD (each product is near point ×\times far point).
  • Secant and tangent from an outside point, with tangent length PTPT:  PT2=PA⋅PB\ PT^2 = PA\cdot PB.

Why the products agree, in the two-chord case: triangles PACPAC and PDBPDB have equal vertical angles at PP and equal inscribed angles ∠A=∠D\angle A = \angle D (both stand on arc CBCB), so they are similar by AA. Then PAPD=PCPB\tfrac{PA}{PD} = \tfrac{PC}{PB}, which cross-multiplies to PA⋅PB=PC⋅PDPA \cdot PB = PC \cdot PD. The outside cases use the same similar-triangle argument, and the tangent is the limiting secant whose two intersection points have merged into one.

Example
Power of a point: two chords
4638ABCDP
Drawn to scale: PA ⋅\cdot PB = 4 ⋅\cdot 6 and PC ⋅\cdot PD = 3 ⋅\cdot 8 are both 24.

Chords AB‾\overline{AB} and CD‾\overline{CD} cross at PP inside the circle. If PA=4PA = 4, PB=6PB = 6, and PC=3PC = 3, find PDPD:

PA⋅PB=PC⋅PD  ⇒  4⋅6=3⋅PD  ⇒  PD=243=8.PA\cdot PB = PC\cdot PD \;\Rightarrow\; 4\cdot 6 = 3\cdot PD \;\Rightarrow\; PD = \frac{24}{3} = 8 .
Example
Power of a point: secant and tangent
469PABT
Tangent squared equals near ×\times far: 62^2 = 4 ⋅\cdot 9.

From outside point PP, a tangent of length PTPT and a secant hitting the circle at AA (near) then BB (far) are drawn. If PA=4PA = 4 and PB=9PB = 9, find PTPT:

PT2=PA⋅PB=4⋅9=36  ⇒  PT=36=6.PT^2 = PA\cdot PB = 4\cdot 9 = 36 \;\Rightarrow\; PT = \sqrt{36} = 6 .
Side note
Where it goes wrong

Multiplying the outside piece by the inside piece. With PA=4PA = 4 and PB=9PB = 9, the chord inside the circle is AB=5AB = 5, and PT2=4⋅5=20PT^2 = 4 \cdot 5 = 20 is wrong. The rule is near ×\times far, both measured from PP: 4⋅9=364 \cdot 9 = 36, so PT=6PT = 6.

Concept
Area of a segment
44
Shaded: quarter-disc 4π\pi minus the triangle 8.

A segment is the region between a chord and its arc — what is left of a sector after you remove the triangle formed by the two radii. So

segment area=(sector area)−(triangle area)=n360 πr2−(area of the two-radius triangle).\text{segment area} = (\text{sector area}) - (\text{triangle area}) = \frac{n}{360}\,\pi r^2 - (\text{area of the two-radius triangle}).
Example
Area of a segment

Find the area of the segment cut off by a 90∘90^\circ arc in a circle of radius 44. The two radii form a right triangle with legs 44 and 44.

sector=90360 π(4)2=14⋅16π=4π,triangle=12(4)(4)=8,segment=4π−8≈12.56−8=4.56.\begin{aligned} \text{sector} &= \frac{90}{360}\,\pi (4)^2 = \tfrac{1}{4}\cdot 16\pi = 4\pi ,\\ \text{triangle} &= \tfrac{1}{2}(4)(4) = 8 ,\\ \text{segment} &= 4\pi - 8 \approx 12.56 - 8 = 4.56 . \end{aligned}
Concept
Finding the center and radius by completing the square

An equation like x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0 is a circle in disguise. To find its center and radius, complete the square on the xx-terms and the yy-terms separately, adding the same amounts to both sides. Each time, take half the middle coefficient and square it.

Example
Completing the square
xy−3−2−112345678−8−7−6−5−4−3−2−11235
x2^2 + y2−^2 - 6x + 4y −- 12 = 0 is the circle centered at (3, −-2) with radius 5.

Put x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0 into center–radius form. Group the xx's and yy's and move the constant over:

(x2−6x)+(y2+4y)=12.\begin{aligned} (x^2 - 6x) + (y^2 + 4y) &= 12 . \end{aligned}

Half of −6-6 is −3-3, squared is 99; half of 44 is 22, squared is 44. Add 99 and 44 to both sides:

(x2−6x+9)+(y2+4y+4)=12+9+4,(x−3)2+(y+2)2=25.\begin{aligned} (x^2 - 6x + 9) + (y^2 + 4y + 4) &= 12 + 9 + 4 ,\\ (x - 3)^2 + (y + 2)^2 &= 25 . \end{aligned}

So the center is (3,−2)(3, -2) and the radius is r=25=5r = \sqrt{25} = 5.

Tip

One rule, many faces. “Half the arc” powers inscribed angles, tangent–chord angles, and cyclic quadrilaterals. “Near ×\times far” powers all three Power-of-a-Point setups. And “sector minus triangle” gives a segment. Spotting which familiar idea is hiding in a hard problem is most of the battle.

Try it
Quick check

Chords AB‾\overline{AB} and CD‾\overline{CD} cross at PP inside a circle, with PA=5PA = 5, PB=8PB = 8 and PC=4PC = 4. Find PDPD and the length of chord CD‾\overline{CD}.

Show the answer

PA⋅PB=PC⋅PDPA \cdot PB = PC \cdot PD: 5⋅8=4⋅PD5 \cdot 8 = 4 \cdot PD, so PD=10PD = \mathbf{10} and CD=4+10=14CD = 4 + 10 = \mathbf{14}.

Why It Works

Theorem
The reason behind the rule
CABOD
The diameter CD splits the angle at C into two first-case angles.

Inscribed angle theorem, first case (one side through the center): the triangle formed by the center and the chord is isosceles (two radii), so its base angles are equal; the central angle is an exterior angle of that triangle and equals the sum of the two equal base angles — twice the inscribed angle. The other cases add or subtract two such configurations. The angle-in-a-semicircle corollary follows instantly: the arc is 180∘180^\circ, so the inscribed angle is 90∘90^\circ.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Equation of a circle
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2

What it means. All points at distance rr from the centre (h,k)(h,k).

Example. Center (2,−1)(2,-1), radius 33: (x−2)2+(y+1)2=9(x-2)^2+(y+1)^2=9.

Why it works. A circle is by definition the set of points a fixed distance from the centre. Writing that distance with the distance formula gives (x−h)2+(y−k)2=r\sqrt{(x-h)^2+(y-k)^2}=r; squaring both sides removes the root.

Tip. Given x2+y2+Dx+Ey+F=0x^2+y^2+Dx+Ey+F=0, complete the square in xx and in yy to recover the centre and radius.

Tip
Circle measurements
C=2πr,A=πr2,s=θ360∘⋅2πr,Asector=θ360∘⋅πr2C=2\pi r,\quad A=\pi r^{2},\quad s=\frac{\theta}{360^\circ}\cdot 2\pi r,\quad A_{\text{sector}}=\frac{\theta}{360^\circ}\cdot \pi r^{2}

What it means. Arc length and sector area are just fractions of the whole circle.

Example. r=5r=5: C=10πC=10\pi, A=25πA=25\pi; a 90∘90^\circ arc has length 90360⋅10π=2.5π\tfrac{90}{360}\cdot10\pi=2.5\pi.

Why it works. π\pi is defined as the ratio of circumference to diameter, so C=πd=2πrC=\pi d=2\pi r. An arc cut by θ\theta degrees is θ360\tfrac{\theta}{360} of the way round, so it takes that fraction of the circumference — and the same fraction of the area.

Tip. Answers "in terms of π\pi" should keep the symbol: write 12π12\pi, not 37.737.7, unless the problem asks you to round.

Tip
Inscribed angle theorem
inscribed angle=12 central angle on the same arc\text{inscribed angle}=\tfrac12\,\text{central angle on the same arc}

What it means. An angle drawn from the circle is half the angle drawn from the centre on the same arc.

Example. An inscribed angle subtending an 80∘80^\circ arc measures 40∘40^\circ.

Why it works. Draw the radius from the centre to the angle's vertex, creating an isosceles triangle (two radii). Its base angles are equal, and the exterior angle at the centre equals their sum — twice the inscribed angle.

Tip. Any angle inscribed in a semicircle is 90∘90^\circ, since the central angle is the 180∘180^\circ diameter.