GRE Geometry: Circles

 A circle is a labeled by its center point: O means the circle with center point O

Diameter

A line segment, generally denoted by the variable, that connects two points on the circles and passes through the center of the circle

 

Radius:- A line segment, generally denoted by the variable r from the center of the circle to any point on the circle.

Chord: A line Segment joining two points on the segment of a circle is a chord.

Tangent:- A line that touches only one point on the circumference of the circle is called Tangent. A line drawn a tangent to a circle is perpendicular to the radius at the point of contact.

Circumference:- The distance around a circle is called the circumference.

An Arc is a portion of the circumference of a circle. The shorter distance between A an B along the circle is called the minor arc.

The longer distance between A and B is called the major arc.

Arc length:- For an arc with a central angle measuring θ Degree:

Length = ( θ / 360 )×(Circumference)

= (θ / 360 )×( π d )

 

Problem 1) What is the length of arc AC of the circle with center O as shown in the figure?

Sol. Arc length= (θ / 360 ) ( π d )= (60 / 360 ) (π (2))

12 π / 6    = 2 π units.

 

Area of circles: The area of circles is given by the formula A=  π r2

A sector is a portion of the circle that is bounded by two radii and an arc.

In a sector Whose central angle measures by Q degrees.

 

Area of sector= (Q / 360) x (Area of circle)

 

Problem 2) What is the area of sector AOC in the circle with the center O shown?

 

Sol. Area of sector AOC= ( 60 / 360 ) (36π )     = ( 6 π ) units.

 

PROPERTIES:

  1. The perpendicular from the center of a circle to a chord bisects the chord and vice versa.

 

If ∠OCB=90 °, THEN AC=BC

If AC=BC , then ∠OCB= 90°

 

2) Equal chords of a circle are equidistant from the center. Conversely, chords equidistant from the center are always equal.

3) Any two angles in the same segment are equal. Thus ∠ACB  = ∠ADB

 

 

4)The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any other point on the circle.

∠AOB= 2∠ACB

5)The angle subtended by a semicircle is a right angle. Conversely, the arc of a circle subtending a right angle at any point on the circle is a semi-circle.

 

If AB is a diameter, then ∠ACB = 90°

IF ∠ACB = 90° , then AB is a diameter.

6)Tangent drawn from common external point to a circle are equal.

 

Problem 3) A 5 by 12 rectangle is inscribed in a circle. What is the circumference of the circle?

Sol.

 

∠BCD= 90°

BD is a diameter (Because The arc of a circle subtending a right angle at any point on the circle is a semi-circle)

In BCD , BC2 + DC2 = BD2 (by Pythagoras theorem)

BD = 13

Circumference of the circle = 2π (Radius) = 13 π units

 

Problem 4) In the figure shown below, if the radius of a circle with centre  P is three times the radius of circle with centre A, ∠BAC=∠QPR, and the shaded area of the circle with centre A is 3 π SQUARE units, then what is the area of the shaded part of circle with centre P?

 

Sol. Let ∠BAC =∠QPR = Q°

The shaded area of circle with centre at A = 3 π sq. units = ( Q / 360° ) π (AC)2

PQ = 3AC

The shaded area of circle with centre at P

= ( Q / 360° ) π (PQ)2

= ( Q / 360° ) π (3AC)2

= (3π x 9)

= 27π sq. units

 

Problem 5) Two congruent, adjacent circles are cut out of a  16 by 8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?

 

 

 

 

 

 

Sol. For the circles, the diameter of the circle is the same as the width of the rectangle.

Remaining area = Area of rectangle – 2 x area of circle

Area of rectangle = Length x Breadth = 8 x 16 = 128 Sq. units

radius of circle = 4 units

Area of circle = π (4)2      = 16 π Sq. units

Remaining area = (128- 2x(16π) ) = ( 128 – 32 π) sq. units

 

Problem 6) The Figure shows an equilateral triangle, where each vertex is the center of a circle. Each circle has a radius of 20. What is the area of the shaded region?

 

Sol.   Side of equilateral triangle =40

Area of equilateral triangle= √3/4    (40)2

=√3/4 (1600) = (400 √3)

Area of three sectors subtending an angle of 60° = 3 x (60/360) (π) (r2)

=(1/2) π (400) = 200π

Area of shaded Region= Area of equilateral triangle- area of three sectors subtending an angle of 60°

= (400√3  – 200 π ) sq. units

 

Problem 7)

 

 

 

 

 

 

 

 

 

If the diameter of the circle is 36, what is the length of arc ABC?

Sol.

 

 

 

 

 

 

 

 

Diameter =36, radius =18

Note that a minor arc is the “short way around” the circle from one point to another, & a major arc is the “long way around”. Arc is thus the same as major arc AC

∠AOC = 2∠ABC = 2(40°) = (80°)

The angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any other point on the circle.

Length of minor arc AC =

\(=\quad (\frac { 80 }{ 360 } )\times (2\pi r)\\ =\quad (\frac { 4 }{ 18 } )(2\pi )(18)\\ =\quad (8\pi )\)

Circumference of circle =

\(=2\pi (18)=36\pi\)

Length of major Arc AC = Circumference – Length of minor Arc AC =

\(=36\pi -8\pi ={ (28\pi ) }^{ Ans }\)

 

Problem 8)

 

 

 

 

 

 

 

 

AB is not a diameter of the circle.

Quantity A:         The area of the circle

Quantity B:         9

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  Since, a diameter is the longest straight line, you can draw from one point on a circle to another (that is, a diameter is the longest chord in a circle), the actual diameter must be greater than 6.

If the diameter were exactly 6, the radius would be 3, & the area would be:

Area \(={ (3) }^{ 2 }\pi =9\pi\)

However, since the diameter must actually be greater than 6, the area must be greater than \(9\pi \). Therefore, Quantity A is greater.

 

Problem 9) In the following figure, O is the centre of the circle & \(\angle ABO={ 30 }^{ 0 }\). Find \(\angle ACB\).

 

 

 

 

 

 

 

 

 

 

Sol.

 

 

 

 

 

 

 

 

 

 

In \(\triangle \quad ABO,\quad AO=OB\) (radii of circle)

\(\angle OBA=\angle BAO={ 30 }^{ 0 }\)

(Because Angles opposite to equal sides are equal)

\(In\quad \triangle ABO,\quad \angle ABO+\angle BAO+\angle AOB={ 180 }^{ 0 }\\ \angle AOB={ 120 }^{ 0 }\\ \angle AOB=2\angle ADB\)

(Because the angle subtended by an arc at the canter of the circle is twice the angle subtended by the same arc at any other point on the circle)

\({ 120 }^{ 0 }=2\angle ADB,\quad \angle ADB={ 60 }^{ 0 }\) \(\angle ADB+\angle ACB={ 180 }^{ 0 }\\\)

(Opposite angles of cyclic quadrilateral are supplementary)

\(\angle ACB={ 120 }^{ 0 }\\\)

 

Directions for questions 10 to 12: In the figure below, X & Y are circles with centres O & O’ respectively. MAB is a common target. The radii of X & Y are in the ratio 4:3 & OM=28cm.

 

 

 

 

 

 

 

 

 

 

 

 

Problem: 10) What is that ratio of the length of OO’ to that of O’M?

a) 1:4

b) 1:3

c) 3:8

d) 3:4

Sol.

 

 

 

 

 

 

 

 

 

 

 

 

\(\triangle MAO’\sim \triangle MBO\\ (\angle OBM=\angle O’AM={ 90 }^{ 0 }(Radius\quad is\quad always\quad perpendicular\quad to\quad the\quad tangent\quad at\quad the\quad point\quad of\quad contact)\\ \angle BMO=\angle AMO'(Common))\)

The radii of X and Y are in the ratio 4:3.

\(\frac { OB }{ O’A } =\frac { 4 }{ 3 } \\ \frac { OB }{ O’A } =\frac { OM }{ O’M } =\frac { BM }{ AM } \\ \frac { 4 }{ 3 } =\frac { OM }{ O’M } (Subtract\quad 1\quad from\quad both\quad sides)\\ \frac { 4 }{ 3 } -1=\frac { OM }{ O’M } -1=\frac { OM-O’M }{ O’M } =\frac { OO’ }{ O’M } \\ { (\frac { 1 }{ 3 } =\frac { OO’ }{ O’M } ) }^{ Ans }\)

d) is correct.

 

Problem 11) What is the radius of circle with centre O?

a) 2cm

b) 3cm

c) 4cm

d) 5cm

Sol.

\(\frac { OB }{ O’A } =\frac { OM }{ O’M } =\frac { BA }{ AM } \\ \frac { 4 }{ 3 } =\frac { OM }{ O’M } =\frac { 28 }{ O’M } ,\quad O’M=21\\ OO’=OM-O’M=28-21=7cm\\\)

7cm = (radius of circle with centre O) + (radius of circle with centre O’)

Let radius of circle with centre O be \(4x\) and radius of circle with centre O’ be \(3x\).

\(7cm=4x+3x\\ x=1\)

Therefore, radius of circle with centre O = 4cm

c) is correct.

 

Problem 12) The length of AM is

a) \(8\sqrt { 3 }\)

b) 10cm

c) 12cm

d) 14cm

Sol.

Radius of circle X=OB=4cm

OM=28cm

\(In\quad \triangle OBM,\quad \angle OBM={ 90 }^{ 0 }\\ { OB }^{ 2 }+{ BM }^{ 2 }={ OM }^{ 2 }\\ OB=4cm,\quad OM=28cm\\ { 4 }^{ 2 }+{ BM }^{ 2 }={ 28 }^{ 2 },\quad { BM }^{ 2 }=768\\ BM=16\sqrt { 3 } \\ \frac { 4 }{ 3 } =\frac { BM }{ AM } =\frac { 16\sqrt { 3 } }{ AM } \\ AM=12\sqrt { 3 }\)

c) is correct.

 

Problem 13) O is the centre of a circle of radius 5 units. the chord AB subtends an angle of \({ 60 }^{ 0 }\) at the centre. Find the area of the shaded portion (approximate value).

 

 

 

 

 

 

 

 

a) 50 Sq. units

b) 75 Sq. units

c) 88 Sq. units

d) 67 Sq. units

Sol. 

Area of circle = \(\pi { (5) }^{ 2 }=25\pi \quad Sq.units\)

In\quad \triangle AOB,\quad \angle AOB={ 60 }^{ 0 }\\ \angle OAB=\angle OBA

(Angles opposite to equal sides are equal)

\(\angle AOB+\angle OBA+\angle BAO={ 180 }^{ 0 }\\ \angle OAB=\angle OBA={ 60 }^{ 0 }\) \(\triangle AOB\) is an equilateral triangle with side 5 units.

Area of equilateral \(\triangle AOB\) =

\(\frac { \sqrt { 3 } }{ 4 } ({ side }^{ 2 })=\frac { \sqrt { 3 } }{ 4 } { (5) }^{ 2 }=\frac { 25\sqrt { 3 } }{ 4 }\)

Area of shaded portion =

\(25\pi -\frac { 25\sqrt { 3 } }{ 4 } =25(\pi -\frac { \sqrt { 3 } }{ 4 } )=67.67\quad Sq.units\)
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