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Circles

Contents

  1. INTRODUCTION
  2. TERMINOLOGY
    1. CHORD
    2. DIAMETER
    3. TANGENT
    4. SECANT
    5. ARC
    6. CENTRAL ANGLE
    7. ANGLE SUBTENDED BY THE ARC
    8. ANGLES INSCRIBED IN AN ARC
    9. INTERCEPTED ARC
    10. CONCENTRIC CIRCLES
    11. CONGRUENT CIRCLES
    12. CIRCUMFERENCE
    13. AREA
    14. SECTOR
    15. SEGMENT
  3. MORE ABOUT TANGENTS
    1. COMMON TANGENT
  4. MORE ABOUT CHORDS AND ARCS
  5. ELLIPSE

Circles


  1. INTRODUCTION

A circle is a set of all those points in a plane which are equidistant from a given point, called the centre of the circle, in the same plane. The distance of any point on the circle from the centre is known as the radius of the circle.

REMEMBER:

  • A circle has one and only one centre.
  • A circle has infinite number of radii and diameters.

  1. TERMINOLOGY

In the given figure, point O is the centre and OA is the radius of the circle. Generally, radius is denoted by ‘r’.

  1. CHORD

A line segment joining two distinct points on the circle is known as a chord.

For example, in the above figure, XY is a chord.

  1. DIAMETER

The longest chord that can be drawn in a circle is known as the diameter, generally denoted by ‘d’. It always passes through the centre of the circle.

For example, in the given figure, PQ is a diameter.

Length of the diameter is double that of the radius.

Diameter, d = 2r

  1. TANGENT

A line which lies in the plane of a circle and touches the circle at only one point is known as a tangent.

i. e. there is only one point common to a circle and the tangent.

The point at which the tangent touches the circle is called the point of contact or point of tangency.

For example, in the given figure, line m is the tangent. E is the point of contact.

  1. SECANT

A line which passes through two distinct points on a circle is known as a secant.

For example, in the given figure, line n is the secant.

  1. ARC

A secant (or a chord) divides the circle in two parts. Each of the two parts along with the two points common with the secant (or chord) is known as an arc. The smaller one is the minor arc and the larger one is the major arc. An arc is given a three-letter name.

For example, in the given figure, arc XEY is the minor arc and arc XSY is the major arc.

  1. CENTRAL ANGLE

An angle made by two points on the circle at the centre is known as a central angle.

For example, in the given figure, AOQ is the central angle made by arc ASQ.

If we draw chord AQ then we can also say “AOQ is the central angle of the chord AQ”.

REMEMBER:

  • Angular measure (or simply measure) of an arc ASQ is denoted by m(arc ASQ). Also, m(arc ASQ) = mAOQ =
  • Angular measure of a minor arc is always less than 180.
  • Angular measure of the major arc = 360 the angular measure of the corresponding minor arc
  • Angular measure of a semicircle is 180.
  • Angular measure of a circle is 360.

  1. ANGLE SUBTENDED BY THE ARC

The angle subtended by an arc is the angle formed by joining each end point of the arc to a point that lies on the opposite arc. In the figure given below, BAC is an angle subtended by arc BXC. If we join B and C, we also say that BAC is an angle subtended by chord BC.

  1. ANGLES INSCRIBED IN AN ARC

BAC is an angle inscribed in arc BYC.

  1. INTERCEPTED ARC

If an angle and a circle is given, the arcs that lie in the interior of the angle are said to be intercepted arcs. Intercepted arcs are shown.

REMEMBER:

  • Angles subtended by the same arc are congruent.
  • Angles inscribed in the same arc are congruent.
  • An angle inscribed in a semicircle is a right angle.

Example 1:

A pole has to be erected on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. The distance of the pole from one of the gates is:

[IIFT 2008]

(1) 8 metres(2) 8.25 metres

(3) 5 metres(4) None of the above

Solution:

Let P be the pole.

As P lies on the circumference and AB is the diameter, m APB = 90

Now, (x + 7)2 + x2 = 132 = 169

x2 + 7x – 60 = 0

(x + 12)(x – 5) = 0

Solving for x, we get,

x = 5 and x + 7 = 12

Hence, option 3.

  1. CONCENTRIC CIRCLES

Two circles in the same plane are known as concentric circles when they have the same centre.

  1. CONGRUENT CIRCLES

Circles having equal radii are called congruent circles.

  1. CIRCUMFERENCE

The distance around the curved line which forms the circle is known as the circumference of the circle. Circumference is denoted by ‘C

The circumference of a circle is given by,

C = 2r

where r is the radius of the circle

Also, C = d

where d is the diameter of the circle.

is a constant, approximately equal to 3.14 or 22/7.

REMEMBER:

  • Length of arc, l 2r
    where r is the radius of the circle and is the angular measure of the arc.

Example 2:

If the angular measure of an arc of a circle with circumference 100 cm is 90, find the length of the arc of the circle.

Solution:

Circumference = 2r = 100

Length of the arc = /360 2r

Length of the arc = 90/360 100 = 25

Thus, the length of the arc of the circle is 25 cm.

Example 3:

Four points A, B, C and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

[CAT 2005]

(1) (2) 1+

(3) 4/3(4) 5

Solution:

The ant will not go into the circles with centers B and C and radius = 1 m

The minimum distance that the ant has to traverse = the distance of the path A-H-G-D

HG = 1 m

AH = GD = 1/4 Circumference of circle = /2

AH + GD = m

The ant must traverse (1 + ) m.

Hence, option 2.

  1. AREA

The space enclosed by a circle is known as the area of the circle. For a circle with radius r, the area is given by,

A = r2

Example 4:

Find the area of a circle with circumference 100 cm.

Solution:

Circumference = 2r = 100

Radius = 50 cm

Hence, area (A) = r2 = (50)2 = 2500 cm2

Example 5:

The radius of circle is so increased that its circumference increased by 5%. The area of the circle then increases by

[SNAP 2009]

(1) 12.5%(2) 10.25%

(3) 10.5%(4) 11.25%

Solution:

Circumference of a circle= 2r, where r is the radius.

Circumference r

As the circumference increases by 5%, the radius also increases by 5%

New radius = 1.05r

As area (radius)2

New area = (1.05)2 old area

= 1.1025 old area

Percentage increase in area = 10. 25%

Hence, option 2.

  1. SECTOR

The part of a circle that is enclosed by an arc and the two radii joining the end points of that arc to the centre of the circle is known as a sector. If the arc of the sector is a minor arc, the sector is known as a minor sector. If the arc of the sector is a major arc, the sector is known as a major sector.

For example, in the above figure, the given sector is denoted as (O-ASQ). Sector (O-ASQ) is a minor sector and sector (O-AYQ) is a major sector.

If r is the radius of a circle and is the angular measure of the arc, then area of the sector,

If l is the length of the arc of a circle and r is the radius of the circle, then area of the sector,

Perimeter of a sector is given by, Ps = l + 2r

Example 6:

Find the area and the perimeter of a sector whose corresponding arc subtends an angle of 60 at the centre of a circle of radius 7 cm.

Solution:

Area of the sector, As = /360 r2

Length of the arc, l = /360 2r

Perimeter of the sector is given by,

Ps = length of the arc + 2(radius) = l+2r

  1. SEGMENT

A chord divides the circular region in two parts, each part known as a segment.

Minor segment: The segment of a circle corresponding to a minor arc.

Major segment: The segment of a circle corresponding to a major arc.

For example, in the given figure, the segment corresponding to arc XEY is a minor segment and that corresponding to arc XSY is a major segment.

Area of the minor segment (i.e. the segment corresponding to minor arc XEY)

= area of sector (O-XEY) area of ∆XOY

Area of the major segment = area of the circle – area of the corresponding minor segment

Example 7:

Find the area of a segment which subtends an angle equal to 90 at the centre of a circle having radius 1 cm.

Solution:

Area of the segment

Example 8:

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is

[CAT 2005]

(1) /4(2) /2 1

(3) /5 (4) 1

Solution:

Let the two circles with centres P and Q intersect at M and N.

Quadrilateral PMQN is a square.

mMPN = mMQN = 90

The area common to both the circles

= 2(Area of sector P-MN – Area of ∆PMN)

=

Hence, option 2.

  1. MORE ABOUT TANGENTS

  1. COMMON TANGENT

A tangent which is common to two or more circles is known as a common tangent. In the case of two non-intersecting, non-touching circles, there are two direct common tangents (DCT) and two transverse common tangents (TCT). In the figure below, lines l and m are DCTs and lines k and j are TCTs.

The length of the DCT and the TCT can be found by applying Pythagoras theorem.

where d = distance between the centres;

r1 and r2 = radii of the circles

Explanation:

Draw a perpendicular from centre Q to AP.

ABQE is a rectangle.

AB = QE

In right angled ∆PEQ, using Pythagoras theorem we get,

(AB)2 = (QE)2 = (PQ)2 (PE)2

AB = length of direct common tangent (DCT)

PQ = distance between the centres of the two circles = d

PE = difference between the radii = (r1 r2)

Similarly,

Let U and F be the points at which transverse common tangent (TCT) (i.e. line j) touches the circles with centres P and Q respectively. Draw ray PU and drop a perpendicular on ray PU from Q. Let the point of intersection be S.

In right angled ∆PSQ, using Pythagoras theorem we get,

(UF)2 = (SQ)2 = (PQ)2 (PS)2

UF = length of transverse common tangent (TCT)

PQ = distance between the centres of the two circles

PS = sum of the radii = (r1 + r2)

REMEMBER:

  • The length of the direct common tangent is always more than the length of the transverse common tangent.

  • A tangent at any point on a circle is perpendicular to the radius at the point of contact

  • From a point outside a circle, exactly two tangents can be drawn to the circle.
  • A segment joining a given point in the exterior of the circle and the point of contact of the tangent drawn from the given point is the tangent segment.
  • The tangent segments drawn from an exterior point are equal in length.
  • For example, in the figure below, PA and PB are tangent segments to the circle drawn from the point in the exterior of the circle. Also, PA = PB

Example 9:

In the given figure, lines AD, AE and CB are tangents to the circle with centre O. If AD = 4 cm, find the perimeter of ∆ABC.

Solution:

AD = 4 = AE… ( tangents drawn from the same point are equal in length)

Similarly, CF = CD and BF = BE

Perimeter of ∆ABC = (AC + CF) + (AB + BF)

= (AC + CD) + (AB + BE)

= (AD) + (AE) = 4 + 4 = 8 cm

Example 10:

Two tangents are drawn to a circle from an exterior point A; they touch the circle at points B and C, respectively. A third tangent intersects segment AB in P and AC in R, and touches the circle at Q. If AB = 20, then the perimeter of triangle APR is:

[FMS 2010]

(1) 42(2) 40.5

(3) 40(4) not determined by the given information

Solution:

Two tangents drawn from a point to a circle are congruent.

AB = AC, PB = PQ and QR = RC

Also, by symmetry of the figure,

PQ = QR

Let PB = PQ = QR = CR = x

As AB = 20, AP = AB – BP = 20 – x

AR = 20 – x

Perimeter of ΔAPR = AP + PR + AR

= 20 – x + 2x + 20 – x

= 40 units

Hence, option 3.

  • Only one tangent can be drawn from a point on the circle, and no tangent can be drawn from a point inside the circle.
  • When two circles in one plane have one and only one point in common, the two circles are said to be tangent circles or touching circles.

  • Centres of tangent circles and their point of contact lie on the same line.

  • In the case of two tangent circles (touching externally), there are two DCT and only one TCT.

  • In the case of two intersecting (overlapping) circles, there are two DCT and no TCT.

  • In the case of two circles touching internally, there is only one DCT and no TCT.

Example 11:

Find the length of the DCT and the TCT for two circles having radii 7 cm and 8 cm. The distance between the centres of the two circles is 20 cm.

Solution:

Distance between the centres = d = 20 cm; r1 = 7 cm and r2 = 8 cm

Example 12:

A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

[CAT 2004]

(1) (2)

(3) (4)

Solution:

AB = AK = 2

Quadrilateral ABCD is a square.

Let the centre of the smaller circle be O and let its radius be x.

=

Hence, option 4.

  1. MORE ABOUT CHORDS AND ARCS

  • A perpendicular drawn from the centre of a circle to a chord bisects the chord. Conversely, the segment joining the centre of a circle and midpoint of a chord is perpendicular to the chord.

Example 13:

Find the distance between two parallel chords of lengths 6 cm and 8 cm in a circle with radius 5 cm.

Solution:

If we draw a line perpendicular to these two parallel chords passing through the centre of the circle, it will bisect the chords, dividing them into two equal halves measuring 3 cm each and 4 cm each, respectively.

Applying Pythagoras theorem and using r = 5 cm, we can find that the distance of the chords from the centre is 4 cm and 3 cm respectively. Hence, the distance between these two chords will be the sum of 4 and 3, i.e. 7 cm if the chords are on opposite sides of the centre.

If the chords are on the same side of the centre then the distance between the chords is the difference of 4 cm and 3 cm = 1 cm.

Example 14:

What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?

[CAT 2005]

(1) 1 or 7 (2) 2 or 14

(3) 3 or 21 (4) 4 or 28

Solution:

The two chords AB and CD can be on the same side or the opposite sides of the centre O.

Let M and N be the midpoints of AB and CD.

MN is the distance between the two chords.

MB = 12 cm and ND = 16 cm

OM and ON are perpendicular to AB and CD respectively.

ON2 = 202 – 162 (By Pythagoras theorem)

ON = 12 cm

Similarly, OM = 16 cm

Case 1: AB and BC are on the same side of the centre.

MN = OM – ON = 4 cm

Case 2: MN = OM + ON = 28 cm

Hence, option 4.

REMEMBER:

  • Equal chords are equidistant from the centre. Conversely, chords equidistant from the centre are equal in length

  • If two secants of a circle passing through point P outside the circle intersect the circle at points A, B, C and D, then
    PA PB = PC PD
    and,

    Explanation:

    In ΔPAD and ΔPCB
    mPDA = mPBC …(angles subtended by arc AC)
    mDPA = mBPC …(common angle)
    ΔPAD ΔPCB … (A-A test)

    PA PB = PC PD

  • If PT is a tangent to a circle and a secant passing through P intersects the circle at A and B then
    PT2 = PA PB

  • If we draw two chords AB and CD intersecting at a common point P inside the circle, then
    PA PB = PC PD

Example 15:

Find CD and PT in the given figure. It is given that PA = 3 cm, AB = 5 cm and PC = 4 cm.

Solution:

Using the formula, PA PB = PC PD = PT PT

3 (3 + 5) = 4 (4 + CD) = PT2

  • Angles subtended by an arc or a chord at distinct points on the circumference, in the same segment, are equal.

  • Angles subtended by congruent chords at the centre are congruent.
  • Angles subtended by a chord at distinct points on the circumference, in alternate segments, are supplementary.

  • The angle subtended by an arc or a chord, at any point on the circumference, is half of the angle subtended by the same arc or chord at the centre of the circle.

Example 16:

In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If CBE = 65, then what is the value of DEC?

[CAT 2004]

(1) 35(2) 55

(3) 45(4) 25

Solution:

mEBC = 65

mEOC = 130 ( angle subtended by an arc at the centre is twice the angle subtended by the same arc on the circumference)

As AC || ED, mOCE = mDEC

ΔOEC is an isosceles triangle.

mOCE = mOEC = (180 – 130)/2 = 25

mDEC = 25

Hence, option 4.

Example 17:

Find m(arc DA) and m(arc BC) in the given figure.

Solution:

m(arc FB) = 2 mBDF = 20

m(arc AD) = 100

Hence, m(arc BC) = 70

Example 18:

In the figure given below, AB is the chord of a circle with centre O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ACD = y and AOD = x such that x = ky, then the value of k is

[CAT 2003 Leaked Test]

(1) 3 (2) 2

(3) 1(4) None of the above

Solution:

Consider the figure shown above.

In ∆OBC, BC = OB

BOC = BCO = y … (i)

Also, OB = OA = Radius of the circle

OBA is an exterior angle of ∆OBC.

OBA = BOC + BCO = 2y

OAB = OBA = 2y … (ii)

In ∆AOB, AOB = 180 4y

Now, AOD + AOB + BOC = 180

x + 180 4y + y = 180

x = 3y

The value of k = 3

Hence, option 1.

  • The angle between a tangent and a chord at the point of tangency is equal to the angle subtended by the chord in the alternate segment. This is known as the alternate segment theorem.

Example 19:

In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ATC = 30 and ACT = 50, then the angle BOA is:

[CAT 2003 Re-Test]

(1) 100(2) 150

(3) 80(4) Cannot be determined

Solution:

In ∆ACT,

ACT = 50 and ATC = 30,

CAT = 100

Applying the Alternate Segment theorem,

ABC = 50

Since CAT is the external angle of ∆ABC, the sum of ABC and BCA is 100,

BCA = 50

BOA = 100 ( angle subtended by an arc at the centre is twice the angle subtended by the same arc on the circumference)

Hence, option 1.

  • A line joining the centres of two intersecting circles is the perpendicular bisector of the common chord passing through the points of intersection of the two circles.

Example 20:

In the given figure, AB = AC. Find the values of BOC and BDC.

Solution:

mEAB = 60… (vertically opposite angle)

mEAB = mACB = 60… (by alternate segment theorem)

mABC = 60 … (AB = AC in ∆ABC)

mBAC = 60

i.e. ∆ABC is an equilateral triangle.

BAC is an angle subtended by arc BDC.

mBOC = m(arc BDC) = 2 60 = 120

BAC and BDC are the angles subtended by BC in alternate segments, and hence are supplementary.

mBDC = 120

Example 21:

Two identical circles, each with radius 5 cm, intersect each other such that the distance between the two centres is 8 cm. Find the length of their common chord.

Solution:

Congruent chords are equidistant from the centre and the circles are congruent implies that PO = QO = 4

We know that the line joining the two centres is perpendicular bisector of the common chord.

OA = OB = 3 cm (by applying Pythagoras theorem).

Hence, the length of common chord = 6 cm

Example 22:

There is a common chord of 2 circles with radius 15 and 20. The distance between the two centres is 25. The length of the chord is

[CAT 2002]

(1) 48(2) 24

(3) 36(4) 28

Solution:

In ∆ABC,

AB2 + AC2 = BC2

BAC = 90

Let BP = x

PC = 25 x

202 = x2 + AP2...(i)

In ∆APC,

225 = AP2 + (25 – x)2

225 = AP2 + 625 + x2 – 50x ...(ii)

Equating (i) and (ii), we get,

225 = 625 + 400 – 50x

50x = 800

x = 16

AP2 = 202 – 162 = 122

AP = 12

Length of the chord = 2 AP = 24

Hence, option 2.

  1. ELLIPSE

An ellipse is the path traced out by a point such that the sum of its distances from two fixed points is constant. The two fixed points are known as foci of the ellipse.

As shown in the figure, points F1 and F2 are the foci of the given ellipse. The sum of the distances of any point (say P) on the ellipse from the two foci is constant. i.e. r1 + r2 = constant, where r1 and r2 are the distances of point P from F1 and F2.

If the line joining the two foci is extended up to the ellipse, we get line segment CD which is known as the major axis. Line segment AB is the perpendicular bisector of the major axis and is known as the minor axis.

Half of the major axis is known as the semimajor axis (line segment CO or DO) and half of the minor axis is known as the semiminor axis (line segment AO or BO).

If semimajor axis = a and semiminor axis = b, then perimeter of the ellipse is given by,

Pe (a + b)

Area of the ellipse is given by,

Ae = ab

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