Geometry

high importance~4 Q in Tier 137 formulas⚡ 19 shortcuts6 subtopics

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Circles: chords, tangents, secants and cyclic angles

⏱ 4 min read🧩 6 question types🎯 13 practice Q

The idea in one minute

A line from the centre perpendicular to a chord cuts the chord in half. So half-chord, distance from centre and radius form a right triangle. The angle a chord makes at the centre is twice the angle it makes at the circle. A tangent meets the radius at $90^\circ$ , and the two tangents from one outside point are equal. Two circles can share 4, 3, 2, 1 or 0 common tangents, decided by how they sit.

Chord, distance, radius

Drop a perpendicular from the centre to a chord. It lands at the chord's midpoint. So:

\left(\frac{c}{2}\right)^2+d^2=r^2

$c$ = chord, $d$ = its distance from the centre, $r$ = radius.

Radius 15, chord 9 from the centre: half-chord $=\sqrt{225-81}=\sqrt{144}=12$ , chord $=24$ .

Rule: Two of the three $(c/2,\ d,\ r)$ always give the third. Circle questions reuse the same Pythagorean triplets.

Equal chords sit at equal distances from the centre. Longer chords sit closer to the centre. Diameter 10 with chord 8: half-chord $4$ , distance $\sqrt{25-16}=3$ . The triplet 3-4-5 again.

Angles in a circle

Angle at the centre $=$ twice the angle at the circle on the same arc: $\angle BOC=2\angle BAC$ .
Angles in the same segment (standing on the same arc) are equal.
An angle in a semicircle (diameter as one side) is $90^\circ$ .

$\angle BOC=120^\circ$ gives $\angle BAC=60^\circ$ instantly.

Tip: When a chord and a circle-angle appear together, double the circle-angle first. The centre angle tells you where the chord sits.

Tangents from an outside point

The tangent meets the radius at $90^\circ$ at the touch point.
The two tangents from one outside point are equal in length.
Tangent length from distance $d$ of the point and radius $r$ : $\sqrt{d^2-r^2}$ .

Point 13 cm from the centre, radius 5: tangent $=\sqrt{169-25}=12$ cm. The line joining the point to the centre also bisects the angle between the two tangents.

Power of a point

Two one-line rules for secants and chords:

Tangent + secant from one outside point: $(\text{tangent})^2=\text{outside part}\times\text{whole secant}$ .
Two chords crossing inside: the products of their pieces are equal, $a\times b=c\times d$ . Pieces 4 and 6 crossing pieces 2 and 12: $4\times6=2\times12=24$ .

Tangent 10 with a 5 cm outside secant part: whole secant $=\dfrac{100}{5}=20$ , so the inside part is 15.

Watch: 'Whole secant' means outside part plus inside part. Using only the inside part is the classic error.

How many common tangents?

Two circles, radii $r_1>r_2$ , centres $d$ apart:

Position	Tangents
Apart ( $d>r_1+r_2$ )	4
Touching outside ( $d=r_1+r_2$ )	3
Overlapping	2
Touching inside ( $d=r_1-r_2$ )	1
One inside the other	0

Radii 6 and 2, centres 4 apart: $4=6-2$ , touching inside, exactly 1 tangent.

Tip: Compare $d$ with $r_1+r_2$ and $r_1-r_2$ . The countdown 4-3-2-1-0 follows the two checks.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Chord, distance, radius triangle

How to spot it:

A chord's length, the radius, or its distance from the centre is missing.

\left(\frac{c}{2}\right)^2+d^2=r^2

Method

Halve the chord: the centre line hits its midpoint.
Write the right triangle: half-chord and distance as legs, radius as hypotenuse.
Solve for the missing one.
Check for a triplet in the halves.

Why it works:

The perpendicular from the centre bisects the chord, so Pythagoras finishes it.

Try this

A circle of radius 15 cm has a chord 9 cm from the centre. Find the chord's length.

Show solution

Half-chord $=\sqrt{15^2-9^2}$ .
$=\sqrt{225-81}=\sqrt{144}=12$ .
Chord $=24$ cm (9-12-15).

Answer

24 cm

Type 2common2 practice Q

Tangent-secant and crossing chords

How to spot it:

A tangent and a secant from one outside point, or two chords crossing inside, with piece lengths to find.

t^2=a(a+b),\quad ab=cd

Method

Square the tangent.
Divide by the outside part to get the whole secant.
Subtract the outside part for the inside piece.
For crossing chords: set the two products equal.

Why it works:

Along any line through the point, the product of the two stretches to the circle is the same.

Try this

From an outside point P, a tangent of 10 cm and a secant with a 5 cm outside part are drawn. Find the whole secant.

Show solution

$10^2=5\times$ whole.
Whole $=\dfrac{100}{5}$ .
$=20$ cm (inside part 15 cm).

Answer

20 cm

Type 3common2 practice Q

Counting common tangents

How to spot it:

Two circles with given radii and centre distance; 'how many common tangents?'

d\gtrless r_1+r_2,\quad d\gtrless r_1-r_2

Method

Compare $d$ with $r_1+r_2$ .
Compare $d$ with $r_1-r_2$ .
Read the count: apart 4, touch outside 3, overlap 2, touch inside 1, nested 0.

Why it works:

Each way the circles touch removes one tangent from the maximum of four.

Try this

Two circles of radii 6 cm and 2 cm have centres 4 cm apart. How many common tangents do they have?

Show solution

$r_1-r_2=6-2=4$ .
$d=4=r_1-r_2$ : touching inside.
Exactly 1 common tangent.

Answer

Type 4very common2 practice Q

Centre angle vs circle angle

How to spot it:

O is the centre; an angle at the centre or at the circle on the same arc is given; the other is asked.

\angle BOC=2\angle BAC

Method

Check both angles stand on the same chord or arc.
Halve the centre angle for the circle angle, or double the circle angle.
Use $90^\circ$ at once if one side is a diameter.

Why it works:

The centre angle takes the whole arc; a circle angle takes half of it.

Try this

In a circle with centre O, $\angle BOC=120^\circ$ on arc BC. Find $\angle BAC$ at the circle.

Show solution

$\angle BAC=\dfrac{\angle BOC}{2}$ .
$=\dfrac{120^\circ}{2}$ .
$=60^\circ$ .

Answer

60°

Type 5very common

Tangent length from an outside point

How to spot it:

A point outside the circle with its distance from the centre (or a tangent and a radius) given; the tangent length is asked.

PT=\sqrt{d^2-r^2}

Method

Note the distance $d$ of the point from the centre and the radius $r$ .
The radius to the touch point is perpendicular to the tangent.
Tangent $=\sqrt{d^2-r^2}$ ; check for a triplet.

Why it works:

Radius, tangent and the line to the centre form a right triangle with $d$ as hypotenuse.

Try this

P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from P.

Show solution

$PT=\sqrt{13^2-5^2}$ .
$=\sqrt{169-25}=\sqrt{144}$ .
$=12$ cm.

Answer

12 cm

Type 6occasional

Alternate segment theorem

How to spot it:

A tangent touches the circle at one end of a chord; the angle between them is linked to an angle on the far arc.

\angle(\text{tangent},\text{chord})=\angle\text{in alternate segment}

Method

Mark the angle between the tangent and the chord.
Find the angle standing on the same chord from the far side.
The two are equal; no computation needed.

Why it works:

Both angles hold half of the same arc, one through the tangent, one through the chord.

Try this

AT is a tangent at A, and AB is a chord. If $\angle TAB=40^\circ$ , find $\angle ACB$ where C is on the circle on the far side of AB from T.

Show solution

$\\angle TAB$ is a tangent-chord angle at chord AB.
Alternate segment: it equals any angle on the far arc AB.
$\\angle ACB=40^\\circ$ .

Answer

40°

Formula sheet

Chord from distance

\ell=2\sqrt{r^2-d^2}

d = distance from centre to chord.

Equal chords

\ell_1=\ell_2\Rightarrow d_1=d_2

Equal chords sit equally far from the centre.

Centre vs circumference angle

\angle BOC=2\angle BAC

Both angles stand on chord BC.

Tangent length

PT=\sqrt{d^2-r^2}

P is d from the centre of a circle of radius r.

Tangent-secant

PT^2=PA\cdot PB

Tangent squared = outside part times whole secant.

Intersecting chords

PA\cdot PB=PC\cdot PD

Two chords crossing inside the circle.

Alternate segment

\angle(\text{tangent},\ \text{chord})=\angle\text{ in alternate segment}

The angle between a tangent and a chord equals the angle the chord makes on the far side.

Common tangents (transverse / direct)

L_T=\sqrt{d^2-(r_1+r_2)^2},\quad L_D=\sqrt{d^2-(r_1-r_2)^2}

d = distance between centres.

Shortcuts that save time

⚡ The triplet inside the circle

Radius, distance and half-chord are the sides of a right triangle. Radius 13 and distance 5 give half-chord 12.

Example

A chord is 5 cm from the centre of a circle of radius 13 cm. Find the chord.

Show solution

Half-chord $=\sqrt{13^2-5^2}$ .
$=\sqrt{169-25}=\sqrt{144}=12$ .
Chord $=2\times12=24$ cm.

Answer

24 cm

⚡ Tangent-secant: multiply the pieces

Tangent squared equals outside part times the whole secant. Whole = outside + inside.

Example

From P, a tangent of 6 cm and a secant with a 4 cm outside part are drawn. Find the inside (chord) part.

Show solution

$6^2=4\times$ whole.
Whole $=\dfrac{36}{4}=9$ .
Inside part $=9-4=5$ cm.

Answer

5 cm

⚡ Count common tangents from d

Compare the centre distance with $r_1+r_2$ and $r_1-r_2$ and read off 4, 3, 2, 1 or 0.

Example

Two circles of radii 4 cm and 9 cm have centres 13 cm apart. How many common tangents?

Show solution

$r_1+r_2=13$ .
$d=13=r_1+r_2$ : touching outside.
3 common tangents.

Answer

⚡ Chord of the outer circle touching the inner one

Concentric circles: a chord of the big circle that just touches the small one is $2\sqrt{R^2-r^2}$ . The small radius is its distance from the centre.

Example

Two concentric circles have radii 25 cm and 7 cm. Find the chord of the larger that touches the smaller.

Show solution

Half-chord $=\sqrt{25^2-7^2}$ .
$=\sqrt{625-49}=\sqrt{576}=24$ .
Chord $=48$ cm.

Answer

48 cm

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Forgetting to double: $\sqrt{r^2-d^2}$ is only half the chord.

Multiply by 2 for the full chord length.

Mistake 02

Using the inside part instead of the whole secant.

Whole secant $=$ outside part $+$ inside part. Then divide the tangent square by it.

Mistake 03

Mixing up $d=r_1+r_2$ and $d=r_1-r_2$ when counting tangents.

Sum means touching outside (3 tangents); difference means touching inside (1).

Mistake 04

Doubling an angle at the circumference that already sits at the centre.

Centre angles are the big ones. Only the circumference angle doubles to give it.

Mistake 05

Treating the angle in the same segment as the alternate segment angle.

Same segment $=$ equal angles on one arc. Alternate segment pairs a tangent with the far arc.

Quick revision

Read this the night before the exam.

$\left(\dfrac{c}{2}\right)^2+d^2=r^2$ : the perpendicular from the centre halves the chord.
$\angle BOC=2\angle BAC$ ; same-segment angles equal; semicircle angle $90^\circ$ .
Tangent $\perp$ radius; two tangents from one point are equal; length $\sqrt{d^2-r^2}$ .
Tangent $^2$ $=$ outside $\times$ whole secant; crossing chords: $ab=cd$ .
Common tangents: apart 4, touch outside 3, overlap 2, touch inside 1, inside 0.
Alternate segment: tangent-chord angle equals the angle on the far arc.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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