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Circle geometry and circle theorems

A circle is the set of points at a fixed distance from a centre. Circle geometry combines measurement with angle reasoning. The formulae answer questions about lengths and areas; the circle theorems turn geometric structure into equations.

These ideas support A-level work with radians, trigonometric proof, coordinate circles, tangents and normals. The main challenge is rarely algebra. It is recognising which fact a diagram makes available.

Prerequisites

Section titled “Prerequisites”

You should be able to:

  • use angles on a straight line, around a point and in a triangle;
  • recognise isosceles triangles;
  • solve a linear equation;
  • work with fractions and exact multiples of π\pi;
  • use Pythagoras’ theorem and basic trigonometry.

Review exact arithmetic or Pythagoras and right-angled trigonometry if either technique interrupts the geometry.

Language of circles

Section titled “Language of circles”

Let the centre be OO.

  • A radius joins OO to the circumference.
  • A diameter is a chord through OO. It has length 2r2r.
  • A chord joins two points on the circumference.
  • An arc is part of the circumference.
  • A sector is bounded by two radii and an arc.
  • A segment is bounded by a chord and an arc.
  • A tangent touches the circle at exactly one point.
  • A cyclic quadrilateral has all four vertices on one circle.

The minor arc or sector is the smaller part. The major arc or sector is the larger part. A diagram may not be drawn accurately, so use labels and stated facts rather than visual guesses.

Circumference and area

Section titled “Circumference and area”

For radius rr and diameter d=2rd=2r,

C=2πr=πdandA=πr2.\boxed{C=2\pi r=\pi d} \qquad\text{and}\qquad \boxed{A=\pi r^2}.

Circumference is a length, so its units are linear. Area uses square units. The radius is squared in the area formula but not in the circumference formula.

Worked example: work backwards from circumference

Section titled “Worked example: work backwards from circumference”

A circle has circumference 18π18\pi cm. Find its radius and area.

From 2πr=18π2\pi r=18\pi,

r=9 cm.r=9\text{ cm}.

Therefore

A=π(9)2=81π cm2.A=\pi(9)^2=81\pi\text{ cm}^2.

Keeping π\pi exact avoids premature rounding.

Scaling insight

Section titled “Scaling insight”

If the radius is multiplied by a scale factor kk, then

\qquad A\mapsto k^2A.$$ For example, tripling the radius triples the circumference but multiplies the area by $9$. ## Arcs and sectors in degrees An angle of $\theta^\circ$ selects the fraction $\theta/360$ of a full circle. Hence

\boxed{\text{arc length}=\frac{\theta}{360}\times2\pi r}

and and

\boxed{\text{sector area}=\frac{\theta}{360}\times\pi r^2}.

The same fraction appears in both formulae. What changes is the whole being divided: $2\pi r$ for circumference and $\pi r^2$ for area. ### Worked example: minor sector A sector has radius $12$ cm and angle $75^\circ$. Find its arc length and area.

\begin{aligned} \text{arc length} &=\frac{75}{360}\times2\pi(12)\ &=5\pi\text{ cm},\[4pt] \text{sector area} &=\frac{75}{360}\times\pi(12)^2\ &=30\pi\text{ cm}^2. \end{aligned}

A useful reasonableness check is $75/360<1/4$. Both answers should therefore be less than one quarter of the corresponding whole-circle measure. ### Worked example: a major arc A chord subtends $110^\circ$ at the centre of a circle of radius $9$ cm. Find the length of the **major** arc. The major angle is $$360^\circ-110^\circ=250^\circ.$$ Thus

\text{major arc length} =\frac{250}{360}\times2\pi(9) =\frac{25\pi}{2}\text{ cm}.

Using $110^\circ$ would calculate the minor arc instead. ### Worked example: area of a segment A minor segment is cut from a circle of radius $10$ cm by two radii enclosing $60^\circ$. Find its exact area. The segment is the sector with the central triangle removed:

\text{segment area}=\text{sector area}-\text{triangle area}.

The triangle has two sides of length $10$ with included angle $60^\circ$, so

\begin{aligned} \text{segment area} &=\frac{60}{360}\pi(10)^2-\frac12(10)(10)\sin60^\circ\ &=\frac{50\pi}{3}-25\sqrt3\text{ cm}^2. \end{aligned}

This is a typical multi-step question: identify the compound region first, then choose a formula for each part. ## The structural idea behind circle theorems Joining points on the circumference to the centre creates isosceles triangles because all radii are equal. Many circle theorems are consequences of this one observation together with familiar angle facts. When solving a problem: 1. mark the centre and any radii; 2. identify the chord, arc, tangent or cyclic quadrilateral involved; 3. state the theorem before using it; 4. introduce algebra only after establishing the angle relationship. ## Angle at the centre For points $A$, $B$ and $C$ on a circle with centre $O$, where $O$ and $C$ lie on the same side of chord $AB$,

\boxed{\angle AOB=2\angle ACB.}

Both angles must stand on the **same arc** $AB$. This is more precise than saying they use the same two letters. ### Why the theorem is true Draw $OC$. Suppose $$\angle ACO=x,\qquad \angle OCB=y.$$ Triangles $AOC$ and $BOC$ are isosceles. Their angles at the centre are therefore $180^\circ-2x$ and $180^\circ-2y$. Angles around $O$ sum to $360^\circ$, so the remaining central angle is

\begin{aligned} \angle AOB &=360^\circ-(180^\circ-2x)-(180^\circ-2y)\ &=2x+2y\ &=2\angle ACB. \end{aligned}

This proof also explains why equal radii matter. ### Worked example The angle at the centre standing on minor arc $PQ$ is $146^\circ$. Find the angle at the circumference standing on the same arc. $$\angle PRQ=\frac{146^\circ}{2}=73^\circ.$$ If the angle at the circumference stood on the other arc, it would not be this angle. ## Angle in a semicircle A diameter subtends a right angle at every point on the remaining circumference:

\boxed{\angle ACB=90^\circ\quad\text{if }AB\text{ is a diameter}.}

This follows immediately because the angle at the centre on a semicircle is $180^\circ$, and the angle at the circumference is half of it. ### Worked example: combine a theorem with trigonometry $AB$ is a diameter of length $14$ cm and $C$ lies on the circle. Given $AC=9$ cm, find $\angle ABC$. Since $AB$ is a diameter, $$\angle ACB=90^\circ.$$ In the resulting right-angled triangle, $AB$ is the hypotenuse and

\cos\angle BAC=\frac{AC}{AB}=\frac9{14}.

Therefore Therefore

\angle BAC=\cos^{-1}!\left(\frac9{14}\right)\approx50.0^\circ,

so $$\angle ABC\approx40.0^\circ.$$ The theorem creates the right angle; trigonometry then completes the calculation. ## Angles in the same segment Angles at the circumference standing on the same chord and in the same segment are equal:

\boxed{\angle ACB=\angle ADB.}

Each is half the same angle $\angle AOB$ at the centre. The phrase **same segment** matters. Points $C$ and $D$ must lie on the same side of chord $AB$. ### Worked example Angles $\angle AXB$ and $\angle AYB$ stand on chord $AB$ in the same segment. If $$\angle AXB=5x+7^\circ, \qquad \angle AYB=8x-32^\circ,$$ then

5x+7=8x-32.

Hence $3x=39$, so $x=13$ and both angles are $$5(13)+7=72^\circ.$$ ## Cyclic quadrilaterals Opposite angles in a cyclic quadrilateral sum to $180^\circ$:

\boxed{\angle ABC+\angle ADC=180^\circ.}

Equivalently, an exterior angle of a cyclic quadrilateral equals the interior opposite angle. ### Why the theorem is true Opposite angles stand on the two arcs joining the other pair of vertices. Those two arcs make a full turn of $360^\circ$. Each angle at the circumference is half the corresponding angle at the centre, so their sum is $$\frac12\times360^\circ=180^\circ.$$ ### Worked example $ABCD$ is cyclic, with $$\angle ABC=3x+18^\circ, \qquad \angle ADC=5x+2^\circ.$$ Since they are opposite,

3x+18+5x+2=180.

Thus $8x=160$, so $x=20$. The angles are $78^\circ$ and $102^\circ$, whose sum checks correctly. The supplementary condition also has a useful converse: if a quadrilateral has a pair of opposite angles summing to $180^\circ$, then its vertices are concyclic. ## Tangent theorems ### Radius perpendicular to a tangent A tangent is perpendicular to the radius at the point of contact:

\boxed{OT\perp \text{ tangent at }T.}

Therefore any triangle containing the centre, the contact point and a point on the tangent is right-angled at the contact point. ### Equal tangents from one point If tangents from an external point $P$ touch the circle at $A$ and $B$, then

\boxed{PA=PB.}

To see why, join $O$ to $A$, $B$ and $P$. The right-angled triangles $OAP$ and $OBP$ have equal radii $OA=OB$ and common hypotenuse $OP$, so they are congruent. ### Worked example: tangent length A circle has centre $O$ and radius $6$ cm. From point $P$, a tangent touches the circle at $T$. If $OP=10$ cm, find $PT$. Since $OT\perp PT$, triangle $OPT$ is right-angled at $T$. By Pythagoras,

PT^2=OP^2-OT^2=10^2-6^2=64.

As $PT$ is a length, $$PT=8\text{ cm}.$$ ### Angle between two tangents If tangents $PA$ and $PB$ touch a circle with centre $O$, then $OA\perp PA$ and $OB\perp PB$. The angles of quadrilateral $OAPB$ sum to $360^\circ$, giving

\boxed{\angle APB=180^\circ-\angle AOB.}

For example, if $\angle AOB=124^\circ$, then the angle between the tangents is $56^\circ$. ## Alternate segment theorem The angle between a tangent and a chord equals the angle in the alternate segment subtended by that chord:

\boxed{\angle\text{ between tangent }AT\text{ and chord }AB=\angle ACB.}

The comparison angle lies on the opposite side of chord $AB$. A reliable way to use the theorem is to trace the chord from the tangent angle, then find the angle at the circumference whose arms end at the same two endpoints. ### Why the theorem is true Let the angle between the tangent at $A$ and radius $AO$ be $90^\circ$. If the base angle of isosceles triangle $AOB$ is $x$, then $$\angle AOB=180^\circ-2x.$$ The angle at the circumference on chord $AB$ is $$\frac12(180^\circ-2x)=90^\circ-x.$$ The angle between the tangent and chord $AB$ is also $90^\circ-x$. The two angles are equal. ### Worked example: a chain of theorems A tangent at $A$ meets chord $AB$. The angle between the tangent and $AB$ is $38^\circ$. Point $C$ lies on the opposite arc $AB$, and $AC$ is a diameter. Find $\angle ABC$ and $\angle BAC$. By the alternate segment theorem, $$\angle ACB=38^\circ.$$ Since $AC$ is a diameter, $$\angle ABC=90^\circ.$$ Angles in triangle $ABC$ sum to $180^\circ$, so

\angle BAC=180^\circ-90^\circ-38^\circ=52^\circ.

Notice that each conclusion has its own justification. ## Choosing the right theorem Use the visible structure as a trigger: | Structure in the question | Likely fact | | --- | --- | | centre and an angle at the circumference on the same arc | centre angle is twice circumference angle | | diameter and a third point on the circle | angle in a semicircle is $90^\circ$ | | two circumference angles standing on one chord | angles in the same segment are equal | | four vertices on a circle | opposite angles sum to $180^\circ$ | | radius meeting a tangent | perpendicular, so $90^\circ$ | | two tangents from one external point | tangent lengths are equal | | tangent and chord meeting | alternate segment theorem | In a multi-step problem, the first theorem often exposes a triangle whose remaining angles follow from isosceles, straight-line or triangle angle facts. ## Common misconceptions - The angle at the centre is twice the angle at the circumference only when both stand on the same arc. - An angle in a semicircle is $90^\circ$ only when its opposite side is a diameter. - Equal angles in the same segment stand on the same chord and lie on the same side of it. - Opposite angles in a cyclic quadrilateral are supplementary, not equal in general. - A tangent is perpendicular to the radius at the point of contact, not to every radius. - The alternate segment theorem uses a tangent and a chord. It does not apply to an arbitrary line through the circle. - Arc length and sector area use the central angle, not an angle at the circumference. - Calculator answers should not replace exact multiples of $\pi$ unless a decimal accuracy is requested. - A result that looks plausible from the sketch still needs a theorem or angle fact as justification. ## Self-check ### 1. Circumference and area A circle has diameter $16$ cm. Find its exact circumference and area. <details> <summary>Answer</summary> The radius is $8$ cm, so $$C=16\pi\text{ cm}, \qquad A=64\pi\text{ cm}^2.$$ </details> ### 2. Arc and sector A sector has radius $15$ cm and central angle $144^\circ$. Find its exact arc length and area. <details> <summary>Answer</summary> Since $144/360=2/5$,

\text{arc length}=\frac25(30\pi)=12\pi\text{ cm},

\text{sector area}=\frac25(225\pi)=90\pi\text{ cm}^2.

</details> ### 3. Centre and circumference An angle at the circumference is $47^\circ$. Find the angle at the centre standing on the same arc. <details> <summary>Answer</summary> $$2\times47^\circ=94^\circ.$$ </details> ### 4. Same segment Two angles in the same segment are $7x-9^\circ$ and $4x+30^\circ$. Find $x$ and the common angle. <details> <summary>Answer</summary> $$7x-9=4x+30,$$ so $3x=39$ and $x=13$. The common angle is $$7(13)-9=82^\circ.$$ </details> ### 5. Cyclic quadrilateral One angle of a cyclic quadrilateral is $116^\circ$. Find its opposite angle. <details> <summary>Answer</summary> Opposite angles sum to $180^\circ$: $$180^\circ-116^\circ=64^\circ.$$ </details> ### 6. Tangents Tangents $PA$ and $PB$ touch a circle at $A$ and $B$. If $PA=3x+5$ cm and $PB=5x-9$ cm, find $x$ and both tangent lengths. <details> <summary>Answer</summary> Tangents from the same external point are equal: $$3x+5=5x-9.$$ Thus $2x=14$, so $x=7$ and $$PA=PB=26\text{ cm}.$$ </details> ### 7. Diagnose an error A student sees a $68^\circ$ angle at the centre and claims that every angle at the circumference with endpoints on the same chord is $34^\circ$. What must be checked? <details> <summary>Answer</summary> The circumference angle must stand on the same arc as the $68^\circ$ central angle. If it stands on the other arc, the relevant reflex central angle is $$360^\circ-68^\circ=292^\circ,$$ and the circumference angle is $146^\circ$, not $34^\circ$. </details> ### 8. Mixed reasoning $AB$ is a diameter of a circle, and the tangent at $A$ makes an angle of $41^\circ$ with chord $AC$. Find $\angle ABC$ and $\angle ACB$. <details> <summary>Answer</summary> By the alternate segment theorem, $$\angle ABC=41^\circ.$$ Because $AB$ is a diameter, $$\angle ACB=90^\circ.$$ </details> ## Where to go next - Explore how changing $r$ changes $2\pi r$ in the [interactive circumference model](/learn/lab/circle-circumference/). - Convert the degree formulae for arcs and sectors into the cleaner A-level forms in [radians](/learn/trigonometry/radians/). - Use centres, radii and tangents algebraically in [coordinate geometry of circles](/learn/coordinate-geometry/circles/). - Combine circle geometry with the [sine rule, cosine rule and triangle area](/learn/foundations/sine-and-cosine-rules/). - Strengthen theorem justification through [algebraic proof and mathematical reasoning](/learn/foundations/proof-and-reasoning/). Secure circle work has a consistent rhythm: identify the structure, state the relevant fact, calculate carefully, and check that the result fits the geometry.