CBSE - Circles

Fill in the blanks:
(i) The centre of a circle lies in ____________ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a _____________ of the circle.
(iv) An arc is a ___________ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____________ of the circle.
(vi) A circle divides the plane, on which it lies, in _____________ parts.

Answer

(i)   The centre of a circle lies in interior of the circle. (exterior/interior)
(ii)  A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle. (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

Write True or False: Give reasons for your answers.
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs, each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.

Answer

(i) True.
All the line segment from the centre to the circle is of equal length.
(ii) False.
We can draw infinite numbers of equal chords.
(iii) False.
We get major and minor arcs for unequal arcs. So, for equal arcs on circle we can't say it is major arc or minor arc.
(iv) True.
A chord which is twice as long as radius must pass through the centre of the circle and is diameter to the circle.
(v) False.
Sector is the region between the arc and the two radii of the circle.
(vi) True.
A circle can be drawn on the plane.

A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose two circles of equal radius, then both circles will cover each other. Therefore, two circles are congruent if they have equal radius.

Consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths.

In ΔAOB and ΔCO'D,

AB = CD (Chords of same length)

OA = O'C (Radii of congruent circles)

OB = O'D (Radii of congruent circles)

∴ ΔAOB ≅ ΔCO'D (SSS congruence rule)

⇒ ∠ AOB = ∠ CO'D (By CPCT)

Hence, equal chords of congruent circles subtend equal angles at their centres

Let us consider two congruent circles (circles of same radius) with centres as O and O'.

In ΔAOB and ΔCO'D,

∠ AOB = ∠ CO'D (Given)

OA = O'C (Radii of congruent circles)

OB = O'D (Radii of congruent circles)

∴ ΔAOB ≅ ΔCO'D (SAS congruence rule)

⇒ AB = CD (By CPCT)

Hence, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Consider the following pair of circles.

The above circles do not intersect each other at any point. Therefore, they do not have any point in common.

The above circles touch each other only at one point Y. Therefore, there is 1 point in common.

The above circles touch each other at 1 point X only. Therefore, the circles have 1 point in common.

These circles intersect each other at two points G and H. Therefore, the circles have two points in common. It can be observed that there can be a maximum of 2 points in common. Consider the situation in which two congruent circles are superimposed on each other. This situation can be referred to as if we are drawing the circle two times.

Steps of construction:
Step I: A circle is drawn.
Step II: Two chords AB and CD are drawn.
Step III: Perpendicular bisector of the chords AB and CD are drawn.
Step IV: Let these two perpendicular bisector meet at a point. The point of intersection of these two perpendicular bisector is the centre of the circle.

Given,
Two circles which intersect each other at P and Q.
To prove,
OO' is perpendicular bisector of PQ.
Proof,
In ΔPOO' and ΔQOO',
OP = OQ (Radii)
OO' = OO' (Common)
O'P = OQ (Radii)
∴ ΔPOO' ≅ ΔQOO' (SSS congruence condition)
Thus,
∠POO' = ∠QOO' --- (i)
In ΔPOR and ΔQOR,
OP = OQ (Radii)
∠POR = ∠QOR (from i)
OR = OR (Common)
∴ ΔPOR ≅ ΔQOR (SAS congruence condition)
Thus,
∠PRO = ∠QRO
also,
∠PRO + ∠QRO = 180°
⇒ ∠PRO = ∠QRO = 180°/2 = 90°
Hence,
OO' is perpendicular bisector of PQ.

OP = 5cm, PS = 3cm and OS = 4cm.
also, PQ = 2PR
Let RS be x.

In ΔPOR,
OP² = OR² + PR² 
⇒ 5² = (4-x)² + PR² 
⇒ 25 = 16 + x² - 8x +  PR²
⇒ PR² = 9 - x² + 8x --- (i)


In ΔPRS,
PS² = PR² + RS² 
⇒ 3² = PR² + x² 
⇒  PR² = 9 - x² --- (ii)

Equating (i) and (ii),
9 - x² + 8x = 9 - x² 
⇒ 8x = 0
⇒ x = 0
Putting the value of x in (i) we get,
PR² = 9 - 0²
⇒ PR = 3cm
Length of the cord PQ = 2PR = 2×3 = 6cm

Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

Proof,
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND --- (i)
and MB = CN --- (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT --- (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii)  and (iii) we get,
MB - ME = CN - EN
⇒ EB = CE

Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Answer

OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD --- (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC --- (ii)
From (i) and (ii),
AM - BM = MD - MC
⇒ AB = CD

Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA² = OM² + AM² 
⇒ 5² = x² + y² --- (i)
Applying Pythagoras theorem in ΔAMB,
AB² = BM² + AM² 
⇒ 6² = (5-x)² + y² --- (ii)
Subtracting (i) from (ii), we get
36 - 25 = (5-x)² - x² -
⇒ 11 = 25 - 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y² + 49/25 = 25
⇒ y² = 25 - 49/25
⇒ y² = (625 - 49)/25
⇒ y² = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB² = BD² + AD²
⇒ AD² = AB² - BD²  
⇒ AD² = a² - (a/2)²
⇒ AD² = 3a²/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.