Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

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Subtopic 6 of 6·← Quadrilaterals and polygons

Circles: chords, tangents, secants and cyclic angles

Chords: a perpendicular from the centre bisects a chord; chord $=2\sqrt{r^2-d^2}$ . Equal chords are equidistant from the centre (and subtend equal angles at the centre).

Angles: angle at the centre $=$ twice the angle at the circumference on the same arc; angles in the same segment are equal; the angle in a semicircle is $90^\circ$ .

Tangents: tangent $\perp$ radius; the two tangents from an external point are equal; tangent–secant $PT^2=PA\cdot PB$ ; two chords crossing inside give $p_1p_2=q_1q_2$ .

Cyclic quadrilateral: opposite angles supplementary; the exterior angle equals the interior opposite angle.

Common tangents of two circles ( $d$ = centre distance): separate ( $d>r_1+r_2$ ) → 4; externally tangent → 3; intersecting → 2; internally tangent → 1; one inside the other → 0. Direct tangent length $=\sqrt{d^2-(r_1-r_2)^2}$ , transverse $=\sqrt{d^2-(r_1+r_2)^2}$ .

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 13 practice Q

Detailed notes

Chord, distance, right triangle

A perpendicular from the centre to a chord bisects it. So radius $r$ , distance $d$ from centre, and half-chord form a right triangle: $\left(\frac{c}{2}\right)^2 + d^2 = r^2$ Two chords are equal iff they are equidistant from the centre; longer chords sit closer to the centre. Every chord question is: given two of $\left(\frac{c}{2}, d, r\right)$ , get the third — usually a half-size triplet (chord 24, $d$ 9, $r$ 15 → 9-12-15).

Tangent and secant power facts

Tangent ⊥ radius at the point of contact; tangents from an external point are equal.
Tangent–secant: $\text{tangent}^2 = \text{external part} \times \text{whole secant}$ .
Intersecting chords (inside): the products of the parts are equal: $a \times b = c \times d$ . These 'power of a point' identities answer two-step questions in one line each. Micro-example: tangent 10, external secant part 5 → whole secant $= \frac{10^2}{5} = 20$ , internal part 15. The equal-tangents fact also drives the 'two tangents from P, find the perimeter of the small triangle' shells: both tangent lengths are equal, so the perimeter is just twice one tangent plus the chord.

Where the 90 degrees come from

The chord theorem, the tangent-radius perpendicular, and the semicircle angle are all Pythagoras or Thales in costume — when a question mixes a chord with an angle at the circumference, first convert the angle to a centre angle (double it), because the centre angle tells you the arc, and the arc tells you where the chord sits.

Counting common tangents

For two circles with radii $r_1 \ge r_2$ , centre distance $d$ :

Configuration	Common tangents
Separate ( $d > r_1 + r_2$ )	4 (2 direct + 2 transverse)
Externally tangent ( $d = r_1 + r_2$ )	3
Intersecting ($	r_1 - r_2
Internally tangent ( $d = r_1 - r_2$ )	1
One inside the other ( $d < r_1 - r_2$ )	0

Memorise the countdown 4-3-2-1-0 against the configuration — the question is only 'which case?'.

Angles in circles

Angle at the centre $= 2 \times$ angle at the circumference on the same arc: $\angle BOC = 2\angle BAC$ .
Angles in the same segment are equal.
Angle in a semicircle is $90^\circ$ (diameter as one side).
Cyclic quadrilateral: opposite angles supplementary (details in the quad subtopic).
Equal chords subtend equal angles at the centre.

Quick revision

$\left(\frac{c}{2}\right)^2 + d^2 = r^2$ — perpendicular from centre bisects the chord.
Tangent $^2$ = external × whole secant; $a\cdot b = c \cdot d$ for crossing chords.
Common tangents: separate 4, external touch 3, intersecting 2, internal touch 1, nested 0.
$\angle BOC = 2\angle BAC$ ; same segment equal; semicircle $90^\circ$ .
Tangents from an external point are equal; tangent ⊥ radius.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Chord–distance–radius right trianglevery common2 practice Q

How to spot it:

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

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

Halve the chord — the perpendicular from the centre bisects it.
Apply Pythagoras with $r$ as hypotenuse.
Watch for half-size triplets: (9,12,15), (8,15,17), (5,12,13).

Why: the perpendicular-to-chord theorem plus Pythagoras is the whole mechanism; only the missing slot changes.

Example: A circle of radius $15$ cm has a chord $9$ cm from the centre. The length of the chord is:

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

Type 2: Tangent–secant and intersecting chordscommon2 practice Q

How to spot it:

A tangent and a secant from an external point, or two chords crossing inside the circle, with segment lengths given/asked.

t^2 = a \cdot (a + b),\quad a b = c d

External point: square the tangent; equate to external part × WHOLE secant.
Inside crossing: equate the products of the two pairs of parts.
Solve the single resulting equation; answer the exact segment asked.

Why: both are the same 'power of a point' — the product of distances along any line through the point is constant.

Example: From an external point $P$ , a tangent of length $10$ cm and a secant are drawn. If the external part of the secant is $5$ cm, the whole secant is:

$10^2 = 5 \times \text{whole} \Rightarrow$ whole $= 20$ cm (internal part $15$ cm).

Type 3: Counting common tangentscommon2 practice Q

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

Compare $d$ with $r_1 + r_2$ and $r_1 - r_2$ .
Map to the case: $d > r_1{+}r_2$ → 4; equal → 3; between → 2; $d = r_1 - r_2$ → 1; less → 0.

Why: transverse tangents exist only when the circles don't overlap; each threshold (touching outside, overlapping, touching inside) removes tangents one at a time.

Example: Two circles of radii $6$ cm and $2$ cm have centres $4$ cm apart. The number of common tangents is:

$d = 4 = 6 - 2$ : internally tangent → exactly $1$ common tangent.

Type 4: Centre angle vs circumference anglevery common2 practice Q

How to spot it:

$O$ is the centre; an angle at the centre or at the circumference on the same arc is given — find the other, or use a semicircle.

\angle BOC = 2\angle BAC

Confirm both angles subtend the SAME arc (or chord).
Centre angle $= 2 \times$ circumference angle; same-segment angles are equal.
Diameter side → angle in the semicircle is $90^\circ$ instantly.

Why: the inscribed angle holds half the central angle's arc — Thales' result covers the diameter case.

Example: In a circle with centre $O$ , $\angle BOC = 120^\circ$ subtending arc $BC$ . The angle $\angle BAC$ at the circumference is:

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

Formulas

Chord from distance

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

Equal chords

\ell_1=\ell_2\Rightarrow d_1=d_2

Tangent–secant

PT^2=PA\cdot PB

Intersecting chords

PA\cdot PB=PC\cdot PD

Cyclic quadrilateral

A+C=180^\circ,\ \text{ext}=\text{interior opposite}

Alternate segment

\angle\text{tangent-chord}=\angle\text{in alternate segment}

Common tangents (transverse)

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

Common tangents (direct)

L_D=\sqrt{d^2-(r_1-r_2)^2}

Shortcut tricks

⚡ Triplet inside a circle

$r$ - $d$ - $\frac{\ell}{2}$ is a right triangle: radius 13 and distance 5 → half-chord 12 → chord 24. Circles reuse the same triplets as Pythagoras questions.

Example: A chord is at distance 5 cm from the centre of a circle of radius 13 cm. Find its length.

Half-chord $=\sqrt{169-25}=12$ ; chord $=24$ cm.

⚡ Tangent–secant: multiply the pieces

Tangent squared = external secant × whole secant. Read the "whole" as external + internal part.

Example: From P, a tangent of length 6 cm and a secant whose external part is 4 cm are drawn. Find the internal (chord) part.

$6^2=4\times\text{whole}\Rightarrow$ whole $=9$ , so the chord inside is $9-4=5$ cm.

⚡ Count common tangents from $d$ vs $r_1, r_2$

Compare $d$ with $r_1+r_2$ and $|r_1-r_2|$ : outside → 4, touching outside → 3, overlapping → 2, touching inside → 1, contained → 0.

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

$d=13=r_1+r_2$ → externally tangent → 3 common tangents.

⚡ Concentric chord trick

A chord of the outer circle tangent to the inner circle has length $2\sqrt{R^2-r^2}$ — the inner radius is the distance from the centre to the chord.

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

$2\sqrt{625-49}=2\times24=48$ cm.

Where students lose marks

Forgetting to double the half-chord: $\sqrt{r^2-d^2}$ is only half the chord.
Tangent–secant: using the internal part instead of the whole secant length.
$d=r_1+r_2$ vs $d=|r_1-r_2|$ confusion changes the tangent count completely.
Angle in the same segment vs angle in the alternate segment mixed up.

Practice sets — 13 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 13 questions

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