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Year 10 Maths Advanced Circle Geometry

Circle Facts

20 practice questions 1 video lesson Theory + worked examples

Circle facts start with the language of circles, such as chord, segment, sector, arc and tangent, and the chord properties: equal chords sit at equal distances from the centre and make equal angles there, and the perpendicular from the centre to a chord bisects it. In NSW Year 10 Advanced maths these facts are used with Pythagoras' theorem and proved (NESA outcome MA5‑CIR‑P‑01).

You will name parts of a circle, use equal chords and equal angles at the centre, and find the distance from the centre to a chord, using Pythagoras' theorem with half the chord. You will find the distance between two parallel chords on the same or opposite sides of the centre, and find the depth of water in a pipe. Harder questions prove that equal chords are the same distance from the centre. It is part of Year 10 Advanced Circle Geometry.

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Theory

A chord joins two points on a circle. Equal chords make equal angles at the centre and sit the same distance from the centre, and the perpendicular from the centre bisects a chord. This Year 10 Advanced maths page names the parts of a circle and uses these facts with Pythagoras' theorem, algebra and congruence proofs.

A radius joins the centre to a point on the circle; every radius has the same length. A chord joins two points on the circle. A diameter is a chord through the centre, two radii long. An arc is part of the circle itself. A chord splits the circle into a minor arc and a major arc.

A sector is the region between two radii and an arc, like a slice of pizza. A segment is the region between a chord and an arc. A tangent touches the circle at one point, and a secant is a line that cuts the circle at two points.

A chord \(AB\) subtends the angle \(\angle AOB\) at the centre \(O\). The distance from the centre to a chord is measured along the perpendicular.

Two radii and a chord always make an isosceles triangle, because \(OA = OB\). Most circle problems start by spotting one.

It comes up if you are heading for Advanced or Extension maths in Years 11 and 12, where you prove the results as well as use them.

Parts of a circleA circle with centre O. A shaded sector is bounded by two radii and an arc. A horizontal diameter passes through O. A chord near the bottom cuts off a shaded segment. An arc on the upper left is highlighted. O sector radius diameter chord segment arc
A chord joins two points on the circle; a diameter is a chord through the centre. A sector is cut off by two radii, a segment by a chord.
Perpendicular from the centre to a chordA circle with centre O and chord AB. A dashed perpendicular OM meets AB at its midpoint M, so AM equals MB. Triangle OMB is right-angled at M with hypotenuse the radius r, one side the distance d and the other side half the chord. O A B M r d ½ chord
The perpendicular from the centre bisects the chord. Then \(r^2 = d^2 + (\text{half chord})^2\).

For a chord at distance \(d\) from the centre of a circle of radius \(r\):

\[r^2 = d^2 + \left(\dfrac{\text{chord}}{2}\right)^2\]
r2=d2+(chord2)2

If the chord subtends \(\theta\) at the centre, the perpendicular also halves that angle, so

\[\dfrac{\text{chord}}{2} = r\sin\dfrac{\theta}{2} \qquad d = r\cos\dfrac{\theta}{2}\]
chord2=rsinθ2
Chord properties. (1) Equal chords subtend equal angles at the centre, and chords that subtend equal angles at the centre are equal. (2) Equal chords are equidistant from the centre, and chords equidistant from the centre are equal. (3) The perpendicular from the centre to a chord bisects the chord. (4) The line from the centre to the midpoint of a chord is perpendicular to it. (5) The perpendicular bisector of any chord passes through the centre.
Two circles that cross. The line joining their centres is the perpendicular bisector of the common chord. Prove it with two pairs of congruent triangles: SSS first (radii, radii, common side), then SAS.
Angles at a point. The angle and the reflex angle at the centre add to \(360^\circ\). A reflex \(\angle AOB = 250^\circ\) means the non-reflex angle is \(110^\circ\).

How to solve a chord problem

  1. Mark every radius as equal; join the centre to the ends of the chord if the radii are not drawn.
  2. Draw the perpendicular from the centre to the chord. It halves the chord and makes a right-angled triangle with the radius as hypotenuse.
  3. Apply Pythagoras' theorem for lengths, trigonometry when an angle at the centre is given, or the angle sum of the isosceles triangle for angles.
  4. State a reason for every step: "radii", "equal chords subtend equal angles at the centre", "angle sum of a triangle".

Problem types

  1. Two expressions for one quantity: if the chords are equal, the expressions for their angles at the centre (or their distances) are equal. Solve the equation, then substitute back.
  2. Parallel chords: find each distance from the centre. Subtract if they are on the same side, add if they are on opposite sides (Example 2).
  3. Depth of liquid in a pipe: the surface is a chord. Depth \(= r - d\) if the surface is below the centre, and \(r + d\) if it is above the centre.
  4. Radius unknown: write the distance in terms of \(r\), e.g. \(d = r - 3\), then \(r^2 = (r-3)^2 + (\text{half chord})^2\). The \(r^2\) terms cancel, leaving a linear equation.
  5. Proofs: name the two triangles, list three matching facts with reasons, state the congruence test (SSS, SAS, AAS or RHS), then the conclusion ("matching sides of congruent triangles").
Example 1 — Radius, then depth
A chord \(AB\) of a circle with centre \(O\) is \(18\) cm long and is \(12\) cm from the centre.
(i) Find the radius.
(ii) The chord cuts off a minor segment. Find its greatest depth \(h\).
Solution

Draw \(OM \perp AB\). The perpendicular from the centre bisects the chord, so \(MB = 9\) cm.

Chord 18 cm long, 12 cm from the centreA circle with centre O and a horizontal chord AB below the centre. The dashed perpendicular OM is 12 cm and MB is 9 cm. The radius OB is r. The short red line h runs from M straight down to the circle. O A B M 12 9 r h

(i) Pythagoras' theorem in \(\triangle OMB\):

\(r^2\)\(=\)\(12^2 + 9^2\)
\(\)\(=\)\(144 + 81 = 225\)
\(r\)\(=\)\(\mathbf{15}\text{ cm}\)
r2=122+92=225

(ii) The deepest point is where \(OM\) extended meets the circle. That point is one radius from \(O\):

\(h\)\(=\)\(r - OM\)
\(\)\(=\)\(15 - 12 = \mathbf{3}\text{ cm}\)
h=15−12=3
Example 2 — Two parallel chords
A circle has radius \(20\) cm. Two parallel chords are \(32\) cm and \(24\) cm long. Find the distance between the chords if they are
(i) on the same side of the centre
(ii) on opposite sides of the centre.
Solution

Find each chord's distance from the centre, using half of each chord.

\(d_1^2\)\(=\)\(20^2 - 16^2 = 144\)
\(d_1\)\(=\)\(12\text{ cm}\)
\(d_2^2\)\(=\)\(20^2 - 12^2 = 256\)
\(d_2\)\(=\)\(16\text{ cm}\)
d1=12,d2=16
Parallel chords on the same side and on opposite sides of the centreTwo circles of radius 20 cm. In each, a 32 cm chord is 12 cm from the centre and a 24 cm chord is 16 cm from the centre. On the left both chords are below the centre; on the right they are on opposite sides. A red line marks the distance between the chords. O 12 16 Same side O 12 16 Opposite sides

The longer chord is closer to the centre.

\(\text{(i)}\ \ 16 - 12\)\(=\)\(\mathbf{4}\text{ cm}\)
\(\text{(ii)}\ \ 16 + 12\)\(=\)\(\mathbf{28}\text{ cm}\)
16−12=4,16+12=28
Example 3 — Reasons in a chain
In a circle with centre \(O\), the chords \(AB\) and \(CD\) are equal and \(\angle AOB = 104^\circ\). Find \(\angle OCD\), giving reasons.
Solution
Equal chords AB and CD with radii drawnA circle with centre O. Chords AB and CD are marked equal. Radii OA, OB, OC and OD are drawn. The angle AOB is 104 degrees and the angle OCD is x. 104° x O A B C D

Equal chords subtend equal angles at the centre, so \(\angle COD = 104^\circ\).

\(OC = OD\) (radii), so \(\triangle OCD\) is isosceles and \(\angle OCD = \angle ODC = x\). Angle sum of \(\triangle OCD\):

\(2x + 104^\circ\)\(=\)\(180^\circ\)
\(2x\)\(=\)\(76^\circ\)
\(x\)\(=\)\(38^\circ\)
2x+104=180,x=38

\(\therefore \angle OCD = \mathbf{38^\circ}\)

Example 4 — Prove a chord property
\(AB\) is a chord of a circle with centre \(O\), and \(OM \perp AB\) with \(M\) on \(AB\). Prove that \(AM = MB\).
Solution
Perpendicular OM from the centre to chord ABA circle with centre O and chord AB. Radii OA and OB are drawn. A red line OM is perpendicular to AB at M. O A B M

Join \(OA\) and \(OB\). In \(\triangle OMA\) and \(\triangle OMB\):

\(\angle OMA\)\(=\)\(\angle OMB = 90^\circ\ \text{(given)}\)
\(OA\)\(=\)\(OB\ \text{(radii)}\)
\(OM\)\(\)\(\text{is common}\)
OA=OB,OMcommon

\(\therefore \triangle OMA \equiv \triangle OMB\) (RHS).

\(\therefore AM = MB\) (matching sides of congruent triangles), so the perpendicular from the centre bisects the chord.

Common pitfalls

Use half the chord. The right-angled triangle has the half chord as one side. Using the whole chord is the most common slip; halve first.
Watch the hypotenuse. The radius is always the hypotenuse, so \(d^2 = r^2 - (\text{half chord})^2\) when you want the distance. Adding instead of subtracting gives a distance longer than the radius.
Same side or opposite sides? Parallel chords on the same side of the centre: subtract the distances. On opposite sides: add them. Sketch both chords before deciding.
A diameter is a special chord. Every diameter is a chord, but a chord is a diameter only if it passes through the centre. A segment of a circle is a region, not a line.

Frequently asked questions

What is the difference between a chord and a diameter?

A chord is any line segment joining two points on a circle. A diameter is a chord that passes through the centre, so it is the longest chord and is twice the radius.

What is the difference between a sector and a segment of a circle?

A sector is bounded by two radii and an arc, like a slice of pizza. A segment is bounded by a chord and an arc, like the piece left when you cut straight across the circle.

How do you find the distance from the centre of a circle to a chord?

Draw the perpendicular from the centre to the chord. It bisects the chord, giving a right-angled triangle with the radius as hypotenuse, so \(d^2 = r^2 - (\text{half chord})^2\).

Are equal chords the same distance from the centre?

Yes. Equal chords are equidistant from the centre, and they subtend equal angles at the centre. The converses are also true.

How do you prove that the perpendicular from the centre bisects a chord?

Join the centre to both ends of the chord. The two triangles formed have a right angle, equal hypotenuses (radii) and a common side, so they are congruent by RHS. Matching sides then show the two halves of the chord are equal.

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