Jugami Geometry

Jugami › Geometry › Circles

Circles

Every point the same distance from the center. That one rule hides a surprising number of theorems.

This unit covers the parts of a circle, circumference and area, arcs and central angles, inscribed angles (including the right angle in a semicircle), why a tangent meets the radius at 90°, and how to find arc lengths and sector areas.

A circle with center O. Points A, B and P lie on the circle. The central angle AOB is 110 degrees and the inscribed angle APB, which opens onto the same arc AB, is 55 degrees, exactly half110°55°ABPO

Parts of a circle

A circle with center O showing its parts: a radius from O to R on the circle, a diameter through O, a chord across the top, a highlighted arc on the lower left, and a tangent line touching the circle at one point at the bottomOradius ORdiameterchordarctangentR
The main parts of a circle with center O.

A circle is the set of all points in a plane at the same distance from one fixed point, the center. We name a circle by its center: “circle O.”

  • A radius joins the center to a point on the circle. All radii of a circle are equal, and we call their length r.
  • A chord joins two points on the circle.
  • A diameter is a chord through the center. Its length is d = 2r, and it is the longest chord.
  • An arc is a piece of the circle between two points.
  • A tangent is a line that touches the circle at exactly one point. A secant is a line that cuts through it at two points.

Circumference and area

For every circle, the distance around (the circumference) divided by the diameter is the same number, called π (pi). It is about 3.14159, and its digits never end or repeat.

  • Circumference: C = πd = 2πr
  • Area: A = πr²

Why is the area πr²? Cut a circle into many thin wedges, like pizza slices, and lay them head to tail. They form something very close to a rectangle whose height is r and whose width is half the circumference, πr. Its area is r · πr = πr².

Example: a circle with radius 5 has circumference 10π ≈ 31.4 and area 25π ≈ 78.5. Answers “in terms of π” keep π as a symbol, which is exact.

Arcs and central angles

A circle with center O and a central angle AOB of 100 degrees. The minor arc AB, highlighted, also measures 100 degrees. The major arc ADB, going the long way round through D, measures 360 minus 100, or 260 degrees100°ABDO∠AOB = 100°minor arc AB = 100°major arc ADB = 260°
A central angle and the arc it cuts off have the same measure.

A central angle has its vertex at the center. The arc inside it is its intercepted arc, and we measure arcs in degrees: an arc’s measure equals its central angle. A whole circle is 360°.

  • A minor arc is less than 180°. We name it with two letters, like arc AB.
  • A major arc is more than 180°. We name it with three letters, like arc ADB, so it is clear which way round we go. Its measure is 360° minus the minor arc.
  • A semicircle is exactly 180°, cut off by a diameter.

An arc’s degree measure is not its length. A 100° arc on a small coin and a 100° arc on a Ferris wheel have the same measure but very different lengths. Length comes later on this page.

Inscribed angles and the semicircle

Left: two inscribed angles, APB and AQB, both open onto the same arc AB, so they are equal. Right: AB is a diameter of a circle with center O, and C is on the circle, so angle ACB is a right angleABPQ∠APB = ∠AQBABCO∠ACB = 90°
Left: angles on the same arc are equal. Right: an angle in a semicircle is a right angle.

An inscribed angle has its vertex on the circle and its sides along two chords. In the figure at the top of the page, ∠APB is inscribed and opens onto arc AB.

Inscribed angle theorem: an inscribed angle is half the central angle on the same arc. So ∠APB = ½ ∠AOB = ½ (arc AB). At the top of the page, ∠AOB = 110° and ∠APB = 55°.

Why it works, in the simplest case: suppose the chord PA passes through O. Then OP = OB (both radii), so △OPB is isosceles and its base angles at P and B are equal, say x each. The central angle ∠AOB is an exterior angle of that triangle, so it equals x + x = 2x. Other positions of P split into two cases like this one, added or subtracted.

Two quick consequences:

  • Same arc, same angle. Every inscribed angle opening onto arc AB from the same side has the same size, because each is half of the same central angle.
  • Angle in a semicircle. If AB is a diameter, the central angle is a straight angle, 180°. Any inscribed angle ACB on it is ½ · 180° = 90°. This is often called Thales’ theorem.

Explorer: half the central angle

Drag point P around the circle (or tab to it and use the arrow keys; hold Shift for bigger steps). Move the slider to change the central angle AOB. Wherever P goes on the far arc, the inscribed angle APB stays half of AOB. Slide all the way to 180° for the semicircle case.

Interactive figure: a circle with center O, fixed points A and B, the orange central angle AOB with its arc, and a draggable point P with the blue inscribed angle APB

Tangents meet the radius at a right angle

A circle with center O and radius OT of 3. A tangent line touches the circle at T and is perpendicular to the radius there. P is on the tangent, 4 from T, and OP is 5, since 3 squared plus 4 squared is 5 squared345TPOtangent
OT ⊥ PT, so △OTP is a right triangle: 3² + 4² = 5².

Tangent theorem: a tangent line is perpendicular to the radius drawn to the point where it touches. In the figure, OT ⊥ PT.

Why: every point of the tangent except T lies outside the circle, so it is farther from O than T is. That makes OT the shortest segment from O to the line, and the shortest segment from a point to a line is the perpendicular one.

This turns many circle problems into Pythagorean theorem problems. If the radius is 3 and P is 5 from the center, the tangent length is PT = √(5² − 3²) = 4. The converse is also true: a line perpendicular to a radius at its outer end is a tangent.

A bonus fact: the two tangent segments from the same outside point to a circle are equal in length.

Arc length and sector area

A 60 degree sector of a circle of radius 6. The arc is one sixth of the circumference, 2 pi, and the sector area is one sixth of the circle's area, 6 pi60°r = 6arc 2πarc = 60/360 · 2π · 6= 2πarea = 60/360 · π · 6²= 6π
A 60° sector is 60/360 = 1/6 of the circle, so it gets 1/6 of the circumference and 1/6 of the area.

A sector is the pizza-slice region between two radii and their arc. Its central angle θ tells you what fraction of the whole circle you have: θ/360.

  • Arc length: (θ/360) · 2πr
  • Sector area: (θ/360) · πr²

In the figure, r = 6 and θ = 60°. The fraction is 60/360 = 1/6, so the arc is 1/6 · 12π = 2π and the sector area is 1/6 · 36π = 6π.

Common mistakes

  • Using the diameter as the radius. In A = πr², halve the diameter first. A circle with diameter 10 has area 25π, not 100π.
  • Doubling instead of halving. The inscribed angle is half the arc; the central angle equals the arc.
  • Mixing up arc measure and arc length. Degrees tell you the fraction of a turn; length also depends on the radius.
  • Forgetting the right angle. When a problem has a tangent, draw the radius to the point of contact and mark 90°.

A short history: Thales, Euclid and Archimedes

Thales of Miletus (about 600 BCE) is traditionally credited with the fact that an angle in a semicircle is a right angle. The claim comes from much later writers. Diogenes Laertius, about 800 years afterward, reports it from the historian Pamphila, and adds that others gave the credit to Pythagoras. So treat the attribution as a tradition, not a record.

Euclid’s Elements (around 300 BCE) proves the main facts of this page in Book III:

  • Proposition 16: the line at right angles to a diameter at its end falls outside the circle (it is a tangent).
  • Proposition 18: the radius to the point of contact is perpendicular to the tangent.
  • Proposition 20: the angle at the center is double the angle at the circumference on the same arc.
  • Proposition 21: angles in the same segment are equal.
  • Proposition 31: the angle in a semicircle is right.

Book XII, Proposition 2 shows that circles are to one another as the squares on their diameters, which means area grows like r².

Archimedes (about 287–212 BCE), in Measurement of a Circle, proved that a circle’s area equals a right triangle with one leg the radius and the other the circumference, which is our ½ · r · 2πr = πr². Using polygons with 96 sides, he showed that π lies between 3 10/71 and 3 1/7. The symbol π itself is much newer: William Jones used it in 1706, and Leonhard Euler made it standard.

Practice: Round Trip

You’ll get five kinds: inscribed and central angles, angles in a semicircle, tangent lengths, arc lengths and sector areas, and circumference and area. For answers in terms of π, type the number in front of π: for 6π type 6. Fractions like 15/2 work too.