Revision: Circles Mathematics SSC (English Medium) 5th Standard Maharashtra State Board

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 interior of a circle is the set of all points inside the circle.

The exterior of a circle is the set of all points outside the circle.

The collection of all points in the plane whose distance from the center is exactly equal to the radius.

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.

Given

O is Centre of circle

PA and PB are tangents

We know that

The tangents drawn from external point to the circle are equal in length.

From point P, PA = PB

⇒ PL + AL = PN + NB …. (i)

From point L & N, AL = LM and MN = NB } …. Substitute in (i)

PL + Lm = PN + MN

⇒ Hence proved.

Let the two circles intersect at points X and Y.

XY is the common chord.

Suppose ‘A’ is a point on the common chord and AM and AN be the tangents drawn A to the circle

We need to show that AM = AN.

In order to prove the above relation, following property will be used.

“Let PT be a tangent to the circle from an external point P and a secant to the circle through

P intersects the circle at points A and B, then 𝑃𝑇² = 𝑃𝐴 × 𝑃𝐵"

Now AM is the tangent and AXY is a secant ∴ 𝐴𝑀² = 𝐴𝑋 × 𝐴𝑌 … . . (𝑖)

AN is a tangent and AXY is a secant ∴ 𝐴𝑁² = 𝐴𝑋 × 𝐴𝑌 … . . (𝑖𝑖)

From (i) and (ii), we have 𝐴𝑀² = 𝐴𝑁²

∴ AM = AN

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 the circles be represented by (i) and (ii) respectively

TQ, TP are tangents to (i)

TP, TR are tangents to (ii)

We know that

The tangents drawn from external point to the circle will be equal in length.

For circle (i), TQ = TP …. (i)

For circle (ii), TP = TR …. (ii)

From (i) and (ii) TQ = TR

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).