Revision: Circle Mathematics (English Medium) ICSE Class 9 CISCE

A circle is a closed curve where all points on the boundary (called the circumference) are at the same distance from a fixed point inside it.

• The fixed point inside the circle is called the center (O)

The radius is a straight line segment that connects the center of the circle to any point on its circumference.

Characteristics:

• Symbol: Usually represented as r

• All radii of a circle have the same length

• A circle has infinite radii (one to every point on the circumference)

• The radius is always half the diameter

• Radius = `"Diameter"/"2"`

 The diameter is a straight line segment that passes through the center of the circle and has both endpoints on the circumference.

Characteristics:

• The diameter passes through the center

• A circle has infinite diameters

• The diameter is the longest possible chord of a circle

• The diameter is twice the radius

• Diameter = 2 × Radius and

A chord is a straight line segment that connects any two points on the circumference of the circle.

Characteristics:

• A circle has infinite chords

• The diameter is the longest chord in any circle

• Chords closer to the centre are longer than chords farther from the center

The line segment, joining any two points on the circumference of the circle, is called a chord. 

Given: A circle with centre O and an external point T from which tangents TP and TQ are drawn to touch the circle at P and Q.

To prove: ∠PTQ = 2∠OPQ.

Proof: Let ∠PTQ = xº.

Then, ∠TQP + ∠TPQ + ∠PTQ = 180º   ...[∵ Sum of the ∠s of a triangle is 180º]

⇒ ∠TQP + ∠TPQ = (180º – x)   ...(i)

We know that the lengths of tangent drawn from an external point to a circle are equal.

So, TP = TQ.

Now, TP = TQ

⇒ ∠TQP = ∠TPQ

`= \frac{1}{2}(180^\text{o} - x)`

`= ( 90^\text{o} - \frac{x}{2})`

∴ ∠OPQ = (∠OPT – ∠TPQ)

`= 90^\text{o} - ( 90^\text{o} - \frac{x}{2})`

`= \frac{x}{2} `

`⇒ ∠OPQ = \frac { 1 }{ 2 } ∠PTQ`

⇒ 2∠OPQ = ∠PTQ

Given: TP and TQ are two tangents of a circle with centre O and P and Q are points of contact.

To prove: ∠PTQ = 2∠OPQ

Suppose ∠PTQ = θ.

Now by theorem, “The lengths of a tangents drawn from an external point to a circle are equal”.

So, TPQ is an isoceles triangle.

Therefore, ∠TPQ = ∠TQP

`= 1/2 (180^circ - θ)`

`= 90^circ - θ/2`

Also by theorem “The tangents at any point of a circle is perpendicular to the radius through the point of contact” ∠OPT = 90°.

Therefore, ∠OPQ = ∠OPT – ∠TPQ

`= 90^@ - (90^@ -  1/2theta)`

`= 1/2 theta`

= `1/2` ∠PTQ

Hence, 2∠OPQ = ∠PTQ.

Suppose CD and AB are two parallel tangents of a circle with center O
Construction: Draw a line parallel to CD passing through O i.e. OP
We know that the radius and tangent are perpendicular at their point of contact.
∠OQC = ∠ORA = 90°
Now, ∠OQC + ∠POQ = 180°          (co-interior angles)
⇒ ∠POQ = 180° - 90° = 90°
Similarly, Now, ∠ORA +∠POR =180°       (co-interior angles)

⇒ ∠POQ = 180° - 90° = 90°
Now,∠POR + ∠POQ = 90° + 90°  =180°
Since, ∠POR and ∠POQare linear pair angles whose sum is 180°
Hence, QR is a straight line passing through center O.

Construction Join PO and OQ
In ΔPOR and ΔQOR
OP = OQ(Radii)
RP = RQ(Tangents from the external point are congruent)
OR = OR (Common)
By SSS congruency, ΔPOR ≅ ΔQOR
∠PRO = ∠QRO(C.P.C.T)

Now,∠PRO+ ∠QRO= ∠PRQ
⇒ 2 ∠PRO = 120°
⇒ ∠PRO = 60°
Now. In ΔPOR
cos 60° `=(PR)/(OR)`

⇒ `1/2 =(PR)/(OR)`
⇒ OR =  2PR
⇒  OR = PR + PR
⇒  OR = PR +RQ

Let ABCD be the quadrilateral circumscribing the circle.
Let E, F, G and H be the points of contact of the quadrilateral to the circle.

To Prove: AB + DC = AD + BC
Proof:
AB = AE + EB
AD = AH + HD
DC = DG + GC
BC = BF + FC

We have:
AE = AH   (Tangents drawn from an external point to the circle are equal.)
Similarly, we have:
BE = BF
DH = DG
CG = CF
Now, we have:
AB + DC = AE + EB + DG + GC
                = AH + BF + DH + CF
               = (AH + DH) + (BF + CF)
                = AD + BC
⇒ AB + DC = AD + BC

Thus, if a quadrilateral is drawn to circumscribe a circle, the sums of opposite sides are equal.
Hence, proved.

We know that tangent segments to a circle from the same external point are congruent
Now, we have
AR = AO, BR = BP and CP = CQ
Now, AB = AC
⇒ AR+ RB= AQ+ QC
⇒ AR + RB = AR + OC
⇒ RB  = QC
⇒  BP = CP
Hence, P bisects BC at P.

We have to prove that

AQ = `1/2` (perimeter of ΔABC)

Perimeter of ΔABC = AB + BC + CA

= AB + BP + PC + CA

= AB + BQ + CR + CA

(∵ Length of tangents from an external point to a circle are equal ∴ BP = BQ and PC = CR)

= AQ + AR  ...(∵ AB + BQ = AQ and CR + CA = AR)

= AQ + AQ  ...(∵ Length of tangents from an external point are equal)

= 2AQ

⇒ AQ = `1/2` (Perimeter of ΔABC)

Hence proved.

Given: PQ is the diameter of the circle. Point P, Q, C are points of contact of the respective tangents.

To prove: ∠AOB = 90°

Construction: Draw seg OC

Proof:

In ∆OPA and ∆OCA,

side OP ≅ side OC   ...[Radii of the same circle]

side OA ≅ side OA   ...[Common side]

side PA ≅ side CA   ...[Tangent segment theorem]

∴ ∆OPA ≅ ∠OCA   ...[SSS test of congruency]

∴ ∠AOP ≅ ∠AOC   ...[C.A.C.T.]

Let m∠AOP = m∠AOC = x   ...(i)

Similarly, we can prove that ∠BOC ≅ ∠BOQ.

Let m∠BOC = m∠BOQ = y   ...(ii)

m∠AOP + m∠AOC + m∠BOC + m∠BOQ = 180°   ...[Linear angles]

∴ x + x + y + y = 180°   ...[From (i) and (ii)]

∴ 2x + 2y = 180°

∴ 2(x + y) = 180°

∴ x + y = 90°   ...(iii)

Now ∠AOB = ∠AOC + ∠BOC

= x + y   ...[From (i) and (ii)]

∴ ∠AOB = ∠AOC + ∠BOC

= x + y    

∴ ∠AOB = 90°   ...[From (iii)] 

In given figure,

`{:(AF = AE),(FB = BD),(EC = DC):}}`   ...(i) [Tangent Segment theorem]

In ▢ODCE,

∠ECD = 90°   ...[∠ACB = 90°, A–E–C, B–D–C]

`{:(∠ODC = 90^circ),(∠OEC = 90^circ):}}`   ...[Tangent theorem]

∴ ∠EOD = 90°  ...[Remaining angle of ▢ODCE]

∴ ▢ODCE is a rectangle.

Also, OE = OD = r   ...[Radii of the same circle]

∴ ▢ODCE is a square   ...`[("A Rectangle is square if it's"),("adjcent sides are congruent")]`

∴ OE = OD = CD = CE = r   ...(ii) [Sides of the square]

Consider R.H.S. = a + b – c

= BC + AC – AB

= (BD + DC) + (AE + EC) – (AF + FB)   ...[B–D–C, A–E–C, A–F–B]

= (FB + r) + (AF + r) – (AF + FB)   ...[From (i) and (ii)]

= FB + r + AF + r – AF – FB

= 2r

= L.H.S.

∴ 2r = a + b – c

Construction: Join OA, OC and OB

We know that the radius and tangent are perpendicular at their point of contact
∴ ∠OCA =  ∠OCB = 90°
Now, In Δ OCA and ΔOCB
∠OCA = ∠OCB = 90°
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
Δ OCA ≅  Δ OCB
∴ CA =CB

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having center O₁
and
ED = EB for the circle having center O₁ 
Now, Adding ED on both sides in EA = EC. we get
EA+ ED  =  EC + ED
⇒  EA + EB = EC + ED
⇒  AB = CD

Since tangents from an exterior point to a circle are equal in length.

∴ AD = AF [Tangents from A]

BD = BE [Tangents from B]

CE = CF [Tangents from C]

Now,

AB = AC

⇒ AB – AD = AC – AD [Subtracting AD from both sides]

⇒ AB – AD = AC – AF [Using (i)]

⇒ BD = CF ⇒ BE = CF [Using (ii)]

⇒ BE = CE [Using (iii)]

Let ABCD be a parallelogram inscribed in a circle.

Now, ∠BAD + ∠BCD

(Opposite angles of a parallelogram are equal.)

And ∠BAD + ∠BCD = 180°

(A pair of opposite angles in a cyclic quadrilateral are supplementary.)

∠BAD + ∠BCD = `(180^circ)/2` = 90°

The other two angles are 90°, and the opposite pair of sides are equal.

∴ ABCD is a rectangle.

Statement:
The angle in a semicircle is a right angle.

Result:

∠ACB = 90^∘

Short Proof (Idea):

• Diameter subtends an angle of 180° at the centre.

• The angle at the circle is half of it.

• Therefore, angle = 90°.

Statement:
Angles in the same segment of a circle are equal.

Result:

∠ACB = ∠ADB

Short Proof (Idea):

• Both angles stand on the same arc.

• The angle at the centre is double each of them.

• Hence, both angles are equal.

Statement:
The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.

Result:

∠AOB = 2∠ACB

Short Proof (Idea):

• Join the centre to the points on the circle.

• Radii form isosceles triangles.

• Angle at centre = sum of angles at the circle.

• Hence, the angle at the centre is double the angle at the circle.

Statement:
If an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a semicircle.

Result:
Arc AB is a semicircle
(or AB is a diameter)

Short Proof (Idea):

• Given angle at the circle ∠ACB = 90°.

• The angle at the centre is double the angle at the circle.

• Therefore, ∠AOB = 2 × 90° = 180°.

• Hence, A, O, and B lie on a straight line, so AB is a diameter.

• Therefore, arc AB is a semicircle.

• Diameter → Longest chord of a circle

• Perpendicular from centre to a chord → Bisects the chord

• Line joining centre to midpoint of a chord → Perpendicular to the chord

• The greater the chord → Nearer to the centre

• The smaller the chord, → farther from the centre

• Equal chords → Equidistant from the centre

• Chords equidistant from centre → Equal in length

• Only one circle passes through three non-collinear points

1. Arc Definition: An arc is a curved portion of a circle's circumference between two points.

2. Two Types: Minor arc (< 180°) and Major arc (> 180°).

3. Semicircle: When the arc angle is exactly 180°, it's called a semicircle.

4. Complete Circle: Minor arc + Major arc = 360° (complete circumference).

Definition: A segment is a region of a circle bounded by a chord and its arc

Two Main Types:

• Minor Segment = smaller piece

• Major Segment = larger piece

Semicircle Special Case:

• Formed when chord = diameter

• Creates two perfectly equal segments

• Each semicircle = half the circle's area