Lesson 35 of 46 · Geometry

Circles: arcs, sectors, inscribed angles and tangents

Work from the radius to circumference, area, arcs and sectors, and use inscribed angles and tangents to turn circle questions into triangle questions.

About 20 minutes

In this lesson

  • Find circumference, area, arc length and sector area from the radius and a central angle.
  • Relate an inscribed angle to the central angle on the same arc.
  • Use the right angle between a tangent and a radius, and the equal radii, to build triangles.

The idea

Everything starts from the radius

A circle is every point at one fixed distance, the radius rr, from its centre. The diameter is 2r2r. Two formulas carry most questions:

C=2πr=πdA=πr2C = 2\pi r = \pi d \qquad A = \pi r^2

Whatever the question gives you (a diameter, a circumference, an area), turn it into the radius first; from there every other measure is one step away. A circumference of 10π10\pi means r=5r = 5 and an area of 25π25\pi. An area of 49π49\pi means r=7r = 7 and a circumference of 14π14\pi.

Answers to circle questions are usually left in terms of π\pi. When you need a size, π≈3.14\pi \approx 3.14, a little more than 3.

Because circumference grows with rr and area with r2r^2, doubling the radius doubles the circumference but multiplies the area by four. All circles are similar to one another, so this is the scaling rule from the similarity lesson at work.

Arcs and sectors are fractions of the whole

A central angle has its vertex at the centre. Its two radii cut off an arc of the circle and a sector (a pie slice) of its area. Both are the same fraction of the whole as the angle is of 360∘360^\circ:

arc length=θ360×2πrsector area=θ360×πr2\text{arc length} = \frac{\theta}{360} \times 2\pi r \qquad \text{sector area} = \frac{\theta}{360} \times \pi r^2

An arc has two different measures, and a question can ask for either. Its degree measure equals the central angle, whatever the size of the circle. Its length is a distance, and depends on the radius. A 60∘60^\circ arc is one sixth of any circle, but it is 2π2\pi long on a circle of radius 6 and π\pi long on a circle of radius 3.

The perimeter of a sector is the arc plus the two radii. Forgetting the radii is a common way to lose an easy mark.

Inscribed angles

An inscribed angle has its vertex on the circle and its sides along two chords. It measures half the central angle that cuts off the same arc. If central angle AOBAOB is 100∘100^\circ, then every inscribed angle ACBACB with CC on the far side of the circle (the longer arc) is 50∘50^\circ, wherever CC sits on that arc.

Two consequences come up often:

  • An angle inscribed in a semicircle is a right angle. If ABAB is a diameter, the central angle is 180∘180^\circ, so any angle ACBACB with CC on the circle is 90∘90^\circ. A triangle drawn inside a circle with one side along a diameter is therefore a right triangle, and the Pythagorean theorem applies.
  • Inscribed angles on the same arc are equal. Points CC and DD on the same side of chord ABAB see it at the same angle.

If CC is on the short arc instead, angle ACBACB cuts off the other arc, 360−100=260∘360 - 100 = 260^\circ, and measures 130∘130^\circ.

Chords, radii and tangents

Any two radii are equal, so a triangle with two vertices on a circle and the third at the centre is isosceles. Look for those triangles whenever a chord and the centre appear together. The line from the centre perpendicular to a chord cuts the chord in half; with the radius as hypotenuse, that is a right triangle.

A tangent touches the circle at exactly one point, and it is perpendicular to the radius drawn to that point. The radius, the tangent segment and the line from the centre to the outside point form a right triangle whose hypotenuse runs from the centre to the outside point.

Describing a circle figure in words

A circle figure is fully described by its centre, its radius or diameter, the points that lie on the circle, and the chords, radii and tangents drawn between them. When you meet one, write down each fact in that form ("OO is the centre, OA=OB=6OA = OB = 6, PTPT is tangent at TT") before you start; most of the work is spotting the equal radii and the right angles those facts imply.

Worked examples

Example 1: Arc length and sector area

A circle has radius 99. A central angle of 80∘80^\circ cuts off an arc and a sector. Find the arc length, the sector area and the sector's perimeter.

  1. The fraction of the circle is 80360=29\frac{80}{360} = \frac{2}{9}.

  2. Circumference =2π×9=18π= 2\pi \times 9 = 18\pi, so the arc length is 29×18π=4π\frac{2}{9} \times 18\pi = 4\pi.

  3. Area =π×92=81π= \pi \times 9^2 = 81\pi, so the sector area is 29×81π=18π\frac{2}{9} \times 81\pi = 18\pi.

  4. The sector's perimeter adds the two radii to the arc: 4π+18≈30.64\pi + 18 \approx 30.6.

Answer

Arc length 4π4\pi, sector area 18π18\pi, sector perimeter 4π+184\pi + 18. Giving 4π4\pi as the perimeter is the tempting slip: it leaves out the two straight edges.

Example 2: Inscribed angle on either arc

Points AA, BB and CC lie on a circle with centre OO. Angle AOBAOB measures 110∘110^\circ. Find angle ACBACB when CC lies on the longer arc ABAB, and when CC lies on the shorter arc ABAB.

  1. With CC on the longer arc, angle ACBACB cuts off the shorter arc, whose central angle is 110∘110^\circ. Half of it: 55∘55^\circ.

  2. With CC on the shorter arc, angle ACBACB cuts off the longer arc, whose central angle is 360−110=250∘360 - 110 = 250^\circ. Half of it: 125∘125^\circ.

  3. Check: AA, C1C_1, BB, C2C_2 form a quadrilateral inscribed in the circle, and 55+125=18055 + 125 = 180, as opposite angles of such a quadrilateral must be.

Answer

55∘55^\circ on the longer arc and 125∘125^\circ on the shorter arc. Taking the inscribed angle equal to the central angle, 110∘110^\circ, is the tempting wrong answer.

Example 3: A tangent from an outside point

A circle has centre OO and radius 55. Point PP lies outside the circle with OP=13OP = 13. Line PTPT is tangent to the circle at TT. Find PTPT and the area of triangle OTPOTP.

  1. The radius OTOT is perpendicular to the tangent at TT, so triangle OTPOTP has a right angle at TT, with hypotenuse OP=13OP = 13.

  2. PT=132−52=169−25=144=12PT = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12, a 5,12,135, 12, 13 triangle.

  3. The legs OTOT and PTPT are base and height: area =12×5×12=30= \frac{1}{2} \times 5 \times 12 = 30.

Answer

PT=12PT = 12 and the area of triangle OTPOTP is 3030. Treating OPOP as a leg and adding squares, 169+25=194\sqrt{169 + 25} = \sqrt{194}, misses that the right angle is at TT.

Common traps

  • Diameter in place of radius

    A question that gives a diameter of 12 needs r=6r = 6 in both formulas. Using 12 in πr2\pi r^2 gives an area four times too large. Write r=r = on your scratch paper before anything else.

  • Inscribed angle equal to the central angle

    The angle at the centre is twice the angle at the circle on the same arc. If a question gives a central angle and asks for an angle with its vertex on the circle, halve it; if it gives the inscribed angle, double it to get the arc.

  • Arc degrees versus arc length

    "The arc measures 60∘60^\circ" and "the arc is 2π2\pi long" are different kinds of statement. Degrees describe the fraction of the circle; length needs the radius as well.

  • Not drawing the radius to the tangent point

    A tangent on its own gives no numbers. The right angle only appears once you draw the radius to the point of contact, and that right angle is usually the key to the question.

Check yourself

Answer each question before you open its answer.

Question 1: An arc from the area

Circles

A circle has area 36π36\pi. What is the length of an arc cut off by a central angle of 60∘60^\circ?

  1. A.π\pi
  2. B.2π2\pi
  3. C.6π6\pi
  4. D.10π10\pi
  5. E.12π12\pi
Show the answer

Answer: B. 2π2\pi

Working

From πr2=36π\pi r^2 = 36\pi, r=6r = 6, so the circumference is 12π12\pi. A 60∘60^\circ arc is 60360=16\frac{60}{360} = \frac{1}{6} of the circle, so its length is 16×12π=2π\frac{1}{6} \times 12\pi = 2\pi.

Why the other options are wrong

  • A. Takes 6 as the diameter, so uses r=3r = 3. An area of 36π36\pi means r2=36r^2 = 36, so the radius itself is 6.
  • C. This is the sector area, 60360×36π\frac{60}{360} \times 36\pi, not the arc length.
  • D. This is the longer arc, the other 300∘300^\circ of the circle.
  • E. This is the whole circumference; the arc is only one sixth of it.

What carries over

Turn any given measure into the radius, then take the angle's fraction of the circumference.

Question 2: Where on the arc?

Quantitative Comparison

Points AA, BB and CC lie on a circle with centre OO. Angle AOBAOB measures 100∘100^\circ, and CC lies somewhere on the longer arc ABAB; its exact position is not given.

Quantity A: the measure of angle ACBACB, in degrees

Quantity B: 5050

  1. A.Quantity A is greater.
  2. B.Quantity B is greater.
  3. C.The two quantities are equal.
  4. D.The relationship cannot be determined from the information given.
Show the answer

Answer: C. The two quantities are equal.

Working

With CC on the longer arc, angle ACBACB cuts off the shorter arc ABAB, whose central angle is 100∘100^\circ. An inscribed angle is half the central angle on the same arc, so angle ACB=50∘ACB = 50^\circ for every such CC.

Why the other options are wrong

  • A. Treats the inscribed angle as equal to the central angle, 100∘100^\circ, or assumes it grows as CC moves. An inscribed angle is half the central angle on the same arc.
  • B. There is no reason for the angle to fall below half the central angle; every inscribed angle on that arc is exactly half.
  • D. Assumes the angle depends on where CC sits. Any point on the longer arc gives the same angle, because they all cut off the same shorter arc.

What carries over

Inscribed angles on the same arc are all equal, so a point's exact position on that arc does not matter.

Question 3: The nearest point of the circle

Circles

Line PTPT is tangent at point TT to a circle with centre OO and radius 88, and PT=15PT = 15. What is the shortest distance from PP to a point on the circle?

  1. A.77
  2. B.99
  3. C.1515
  4. D.1717
  5. E.2525
Show the answer

Answer: B. 99

Working

The radius OTOT is perpendicular to the tangent, so OP=82+152=64+225=289=17OP = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 (an 8,15,178, 15, 17 triangle). The nearest point of the circle lies on segment POPO, one radius from OO, so the distance is 17−8=917 - 8 = 9.

Why the other options are wrong

  • A. Subtracts the radius from the tangent length. The tangent segment does not point at the centre, so the radius cannot be taken off it.
  • C. This is the distance to TT. TT is on the circle, but the point on line POPO is closer.
  • D. This is the distance from PP to the centre, not to the circle.
  • E. This is the distance to the farthest point of the circle, 17+817 + 8.

What carries over

Draw the radius to the point of tangency: the right angle there turns the question into a Pythagorean one.

Key points

  • Turn every given measure into the radius first; C=2πrC = 2\pi r, A=πr2A = \pi r^2.

  • Arcs and sectors are the fraction θ360\frac{\theta}{360} of the circumference and of the area.

  • An inscribed angle is half the central angle on the same arc; an angle in a semicircle is a right angle.

  • A tangent is perpendicular to the radius at the point of contact.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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