NDA Maths · Circles

Inscribed Geometry, Tangents & Segments

The geometry that lives ON the circle: the angle a chord subtends from the circumference, circles that touch the axes, inscribed squares, the tangent–normal relationship, and the areas of the two segments a chord cuts off.

All Circles notes NDA Maths strategy

Why this matters

A small but HARD-leaning pocket (7 PYQs, 4 HARD, in two passage sets plus singles). The marks come from a handful of named facts — the angle in a semicircle is a right angle, the inscribed angle is half the central angle, a tangent is perpendicular to the radius at the point of contact, and the segment-area split — applied to a circle you've already put in centre–radius form. Knowing which fact a question is fishing for is most of the battle.

Concept 1 of 5

Inscribed Angle and the Angle in a Semicircle

Intuition

Stand anywhere on a circle and look at a fixed chord: the angle you see is always the same, and it is exactly half the angle the chord makes at the centre. The special case — a diameter — is seen as a perfect right angle from every point on the circle.

Definition

Two linked facts about angles a chord subtends:

Inscribed-angle theorem: the angle a chord $BC$ subtends at a point $A$ on the circle is half the angle it subtends at the centre: $\angle BAC = \tfrac12\angle BOC$ .
Angle in a semicircle: if $BC$ is a diameter, $\angle BAC = 90^\circ$ for every $A$ on the circle.
The inscribed point $A$ can sit on either arc: on the major arc the angle is $\tfrac12\angle BOC$ ; on the minor arc it is the supplement, $180^\circ-\tfrac12\angle BOC$ . So a single chord can give an angle AND its supplement — both are valid.

Inscribed angle

\angle BAC = \tfrac12\,\angle BOC

Ocentre
Apoint on the circle

Worked example

B=(3,0)

and

C=(0,3)

lie on a circle centred at the origin

O

. If

A

is another point on the circle, what is

\angle BAC

At the centre: $\vec{OB}=(3,0)$ , $\vec{OC}=(0,3)$ are perpendicular, so $\angle BOC=90^\circ$ .
Inscribed angle is half the central angle: $\angle BAC=\tfrac12(90^\circ)=45^\circ$ when $A$ is on the major arc.
If $A$ is on the minor arc, $\angle BAC=180^\circ-45^\circ=135^\circ$ .

Answer:

45^\circ

(major arc) or

135^\circ

(minor arc).

From the bank · past-year question

Example 1CirclesHARD

A triangle

ABC

is inscribed in the circle

x^2+y^2=100

B

and

C

have coordinates

(6,8)

and

(-8,6)

respectively.

What is

\angle BAC

equal to?

[Q61 · Sep · 2024]

Don't forget the supplementary (obtuse) case

The inscribed angle depends on which arc

A

is on:

\tfrac12\angle BOC

on the major arc, its supplement on the minor arc. NDA answer keys frequently list BOTH

\pi/4

and

3\pi/4

. Quoting only the acute value loses the obtuse option.

A is not pinned to one coordinate

If a question asks for "the coordinates of

A

" with only

B,C

given, there are infinitely many valid points on the arc — the answer is a locus, not a single point. Watch for the choice that says "cannot be uniquely determined".

Drill 1 more on inscribed angle and the angle in a semicircle

Concept 2 of 5

Points Where a Circle Touches the Axes

Intuition

When a circle touches both axes, the points of contact sit directly below and beside the centre. The contact point on the x-axis shares the centre's x-coordinate; the one on the y-axis shares its y-coordinate — so the chord joining them is easy to measure.

Definition

A circle with centre $(k,k)$ and radius $|k|$ (so it touches both axes) meets:

the x-axis at $P=(k,0)$ — directly below the centre;
the y-axis at $Q=(0,k)$ — directly beside the centre.
The distance between the two contact points is $PQ=\sqrt{k^2+k^2}=\sqrt2\,|k|$ . More generally, the contact point on an axis drops the perpendicular from the centre onto that axis.

Contact points and PQ

P=(k,0),\;\; Q=(0,k),\;\; PQ=\sqrt2\,|k|

Worked example

The circle

x^2+y^2-6x-6y+9=0

touches both axes at

P

and

Q

. Find

PQ

Complete the square: $(x-3)^2+(y-3)^2=9$ , centre $(3,3)$ , radius $3$ — touches both axes.
Contact points $P=(3,0)$ , $Q=(0,3)$ .
$PQ=\sqrt{3^2+3^2}=3\sqrt2$ .

Answer:

PQ=3\sqrt2

From the bank · past-year question

Example 2CirclesMODERATE

The circle

x^2+y^2-2kx-2ky+k^2=0

touches the

x

-axis at

P

and

y

-axis at

Q

. What is

PQ

equal to?

[Q64 · Sep · 2025]

The contact point shares ONE coordinate with the centre

On the x-axis the contact point is

(k,0)

— same x as the centre, y zero — because the radius to the contact point is vertical. Reading it as

(0,k)

(k,k)

is the usual mix-up.

Concept 3 of 5

A Square Inscribed in a Circle

Intuition

A square inscribed in a circle has its four corners on the circle and its diagonal equal to the diameter. If its sides are parallel to the axes, each vertex sits at the centre offset by r/√2 in both x and y — the corners of a tilted-by-45° box of half-side r/√2.

Definition

For a square inscribed in a circle of centre $(h,k)$ , radius $r$ , with sides parallel to the axes:

The diagonal of the square is the diameter $=2r$ ; the half-diagonal to each vertex is $r$ .
Each vertex is at $\left(h\pm\tfrac{r}{\sqrt2},\,k\pm\tfrac{r}{\sqrt2}\right)$ — the centre offset by $\tfrac{r}{\sqrt2}$ in each direction (since the side is $r\sqrt2$ and the half-side is $\tfrac{r}{\sqrt2}$ ).
The square's side is $r\sqrt2$ and its area is $2r^2$ .

Inscribed-square vertices

\left(h\pm\tfrac{r}{\sqrt2},\; k\pm\tfrac{r}{\sqrt2}\right)

Worked example

A square with sides parallel to the axes is inscribed in

x^2+y^2=4

. Give one vertex.

Centre $(0,0)$ , radius $r=2$ .
Vertices are at $\left(\pm\tfrac{2}{\sqrt2},\pm\tfrac{2}{\sqrt2}\right)=(\pm\sqrt2,\pm\sqrt2)$ .

Answer:

(\sqrt2,\sqrt2)

(and its sign variants).

From the bank · past-year question

Example 3CirclesMODERATE

A square is inscribed in a circle

x^2 + y^2 + 2x + 2y + 1 = 0

and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?

[Q57 · Apr · 2025]

Inscribed vs circumscribed — offset is r/√2, not r

An INSCRIBED square (corners on the circle) has vertices offset

r/\sqrt2

from the centre. A CIRCUMSCRIBED square (sides tangent to the circle) has vertices offset

r

. Mixing the two gives

(h\pm r,k\pm r)

— the wrong, larger square.

Concept 4 of 5

Tangent and Normal at a Point of Contact

Intuition

At any point of a circle the tangent is perpendicular to the radius, so the normal (perpendicular to the tangent) points straight along the radius — through the centre. Extending the normal across the circle therefore lands you at the diametrically opposite point.

Definition

At a point of contact on a circle:

The tangent is perpendicular to the radius drawn to that point.
The normal (perpendicular to the tangent) is the radius produced — it passes through the centre.
Following the normal across the circle reaches the diametrically opposite point: from contact point $T$ and centre $C$ , the far point is $2C-T$ .
If the y-axis touches $x^2+y^2+2gx+2fy+c=0$ , the contact point is $(0,-f)$ ; the other end of that diameter is $(-2g,-f)$ (the normal is horizontal, through the centre $(-g,-f)$ ).

Opposite end of the diameter

T'=2C-T\quad(C=\text{centre},\;T=\text{contact point})

Worked example

The x-axis touches the circle

x^2+y^2-6x-4y+9=0

. Find the point diametrically opposite the point of contact.

Centre $(3,2)$ , radius $\sqrt{9+4-9}=2$ . It touches the x-axis (centre height $2=$ radius) at $T=(3,0)$ .
The normal is the vertical diameter $x=3$ ; the opposite end is $T'=2C-T=(6-3,\,4-0)=(3,4)$ .

Answer:

(3,4)

From the bank · past-year question

Example 4CirclesHARD

If y-axis touches the circle

x^2+y^2+gx+fy+\frac{c}{4}=0

, then the normal at this point intersects the circle at the point

[Q62 · Sep · 2018]

The normal goes through the centre — that's the whole trick

The normal at a circle's point is the radius line, so it always passes through the centre. The far intersection with the circle is the diametrically opposite point,

2C-T

. Trying to solve the normal–circle intersection from scratch wastes time and invites sign errors.

Concept 5 of 5

Areas of the Minor and Major Segments

Intuition

A chord splits a disc into two segments. The smaller one (minor) is a circular sector minus the triangle the chord cuts off; the larger one (major) is the rest of the disc. Find the central angle the chord subtends, compute the sector and triangle, and the two segments follow.

Definition

A chord subtending a central angle $\theta$ (in radians) in a circle of radius $a$ splits the disc into two segments:

Minor segment $=$ sector $-$ triangle $=\dfrac{a^2}{2}(\theta-\sin\theta)$ .
Major segment $=$ whole disc $-$ minor segment $=\pi a^2-\dfrac{a^2}{2}(\theta-\sin\theta)$ .
Find $\theta$ from the perpendicular distance $d$ from the centre to the chord: $\cos\tfrac\theta2=\dfrac{d}{a}$ . For a chord at distance $\dfrac{a}{\sqrt2}$ , $\tfrac\theta2=45^\circ$ , so $\theta=90^\circ=\tfrac\pi2$ .

Minor segment area

A_{\text{minor}} = \tfrac{a^2}{2}\,(\theta - \sin\theta)

aradius
\thetacentral angle (radians)

Worked example

A chord of a circle of radius

a

subtends a right angle at the centre. Find the minor and major segment areas.

$\theta=\tfrac\pi2$ , $\sin\theta=1$ . Minor segment $=\tfrac{a^2}{2}\left(\tfrac\pi2-1\right)=\tfrac{(\pi-2)a^2}{4}$ .
Whole disc $=\pi a^2$ , so major segment $=\pi a^2-\tfrac{(\pi-2)a^2}{4}=\tfrac{(3\pi+2)a^2}{4}$ .

Answer:Minor

=\tfrac{(\pi-2)a^2}{4}

; major

=\tfrac{(3\pi+2)a^2}{4}

From the bank · past-year question

Example 5CirclesHARD

Consider the following for the items that follow: The line

y=x

partitions the circle

(x-a)^{2}+y^{2}=a^{2}

in two segments.

What is the area of minor segment?

[Q59 · Apr · 2023]

Segment = sector − triangle (not sector alone)

The minor SEGMENT is the sector with the triangle cut off:

\tfrac{a^2}{2}(\theta-\sin\theta)

. Forgetting the

-\sin\theta

term reports the SECTOR area instead — a different region. The major segment is then the whole disc minus the minor segment, not 'the big sector'.

Drill 1 more on areas of the minor and major segments

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Inscribed Angle and the Angle in a Semicircle
Inscribed angle
$\angle BAC = \tfrac12\,\angle BOC$
Points Where a Circle Touches the Axes
Contact points and PQ
$P=(k,0),\;\; Q=(0,k),\;\; PQ=\sqrt2\,|k|$
A Square Inscribed in a Circle
Inscribed-square vertices
$\left(h\pm\tfrac{r}{\sqrt2},\; k\pm\tfrac{r}{\sqrt2}\right)$
Tangent and Normal at a Point of Contact
Opposite end of the diameter
$T'=2C-T\quad(C=\text{centre},\;T=\text{contact point})$
Areas of the Minor and Major Segments
Minor segment area
$A_{\text{minor}} = \tfrac{a^2}{2}\,(\theta - \sin\theta)$

Watch out for (6)

Don't forget the supplementary (obtuse) case
→ Inscribed Angle and the Angle in a Semicircle
A is not pinned to one coordinate
→ Inscribed Angle and the Angle in a Semicircle
The contact point shares ONE coordinate with the centre
→ Points Where a Circle Touches the Axes
Inscribed vs circumscribed — offset is r/√2, not r
→ A Square Inscribed in a Circle
The normal goes through the centre — that's the whole trick
→ Tangent and Normal at a Point of Contact
Segment = sector − triangle (not sector alone)
→ Areas of the Minor and Major Segments

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1CirclesHARD

A triangle

ABC

is inscribed in the circle

x^2+y^2=100

B

and

C

have coordinates

(6,8)

and

(-8,6)

respectively.

What are the coordinates of

A

[Q62 · Sep · 2024]

Example 2CirclesMODERATE

Consider the following for the items that follow: The line

y=x

partitions the circle

(x-a)^{2}+y^{2}=a^{2}

in two segments.

What is the area of major segment?

[Q60 · Apr · 2023]

Drill every past-year question on this subtopic

7 questions from the bank — paginated, with cart and Word-export support.

Drill the 7 questions