# Geometry

#### Coordinate Geometry - 2

CIRCLES- PART-1

CIRCLES- PART-1

This fixed point is called the centre and the fixed distance is known as radius of the circle.

Some common terms associated with Circles:

1. Chord of a Circle: A line segment joining any two points on the circle is known as chord of a circle.

2. Diameter: Diameter is a chord that passes through the centre of a circle. Also it the longest chord of a circle. The length of diameter is equal to twice the radius of a circle.

3.  Circumference: The Circumference is the distance around the edge of the circle.

Circumference = π × Diameter

Circumference = 2 × π × Radius

4. Major Arc and Minor Arc: The arc of a circle consists of two points on the circle and all of the points on the circle that lie between those two points. It's like a segment that was wrapped partway around a circle. Usually an arc is measured by the angle it subtends at the centre.

An arc whose measure is less than 180 degrees is called a minor arc. An arc whose measure is greater than 180 degrees is called a major arc. An arc whose measure equals 180 degrees is called a semicircle, since it divides the circle in two.

Sectors and Segments:

A sector of a circle is the region enclosed by the central angle of a circle and the circle itself.

A segment of a circle is the region enclosed by a chord and the arc that the chord defines.

Area of Sector and Length of an Arc

Example 1. A sector of 450 in a circle of radius 8 units is removed . What is the length of the wire required to fence the rest of the field? Take π = 22/7

Solution:

The required length is the major arc AB subtending an angle of 3150 and the straight lines OA and OB.

Length of arc AB = (315/360) * (22/7) * 82

= (7/8) * (22/7) * 64 = 176

Length of OA and OB =8 + 8 = 16

Length of wire required = 176 + 16 = 192 units.

Example 2. From four corners of a square sheet of side 4 cm, four pieces, each in the shape of arc of a circle with radius 2 cm are cut out. The area of the remaining portion is:

1. (8-π) sq. cm.         2. (16-4π) sq. cm.        3. (16-8π) sq. cm.   4. (4-2π) sq. cm.

Solution:

The area of the remaining portion = Area of square – area of 4 quarters

= Area of square – Area of a circle of radius 2 units

= 16 - 4π

Example 3. In the given figure, two circles each with centre D, have radii of 2 and 3. The total area of the shaded regions is 11/27 times the area of the larger circle. What is the measure of angle ADC?

A. 1080          B. 1200            C. 900           D. 1500

Solution:

There are a few interesting properties of circles that we need to study before we move on to solve problems based on circles. One by one we shall understand them and solve a few questions based on them.

Property 1:

As shown in the figure OM is perpendicular to chord AB. So, AM = MB i.e. OM acts as a perpendicular bisector of AB. The length of OM is also the distance of AB from the centre.

Example 4. If a chord of length 16 cm is at a distance of 15 cm from the centre of the circle, then the length of the chord of the same circle which is at a distance of 8 cm from the centre is equal to

1. 10 cm               2. 20 cm               3. 30 cm                 4. 40 cm

Solution:

Example 5 . AB and CD are two parallel chords drawn on two opposite sides of their parallel diameter such that AB = 6cm, CD =8 cm. If the radius of the circle is 5 cm, the distance between the chords in cm, is-

1. 7                     2. 8                      3. 5                    4. 3

Solution :

The two chords are parallel to each other. So, the distance between them is MN which is OM + ON

OM is the perpendicular distance of AB from the centre while ON is the perpendicular distance of CD from the centre.

Consider ? OMB, OB2 = OM2 + MB2

52 = OM2 + 32

Or OM = 4

Similarly in ?OND, applying Pythagoras theorem we get, ON = 3 units

MN = OM + ON = 4+3 = 7 units.

Example 6. AB is a chord which bisects the radius of a circle perpendicularly. What is the length of AB if the area of the circle is 64π ?

A. 4√3               B. 8                     C. 8√3              D. None of these

Solution:

Given that area of circle = 64π

πr2 = 64π  or  r = 8

In ?OAM

OA2 = OM2 + AM2

82 = 42 + x2

x2 = 48  or x = 4√3

AB = 2x = 8√3

Example 7. If a chord of a circle of radius 5 cm is a tangent to a circle of radius 3 cm, both the circles being concentric, then the length of the chord is-

1. 10 cm                2. 12.5 cm              3. 8 cm                4. 7 cm

Solution:

Property 2:

Example 8.  The length of a chord of  a circle is equal to the radius of the circle. The angle which this chord subtends in the major segment of the circle is-

1. 300                2. 450                 3. 600                  4. 900

Solution:

As OA = OB = AB = r (radius)

So, ?OAB is equilateral.

Thus, angle AOB = 600

Arc AB subtends angle of 60 degrees at the centre, so it will subtend an angle of 30 degrees at any point on the major arc AB.

Example 9. AB is a diameter of a circle with centre O. CD is a chord equal to the radius of the circle. AC and BD are produced to meet at P. Then, the measure of angle APB is-

1. 1200             2. 300               3. 600           4. 900

Solution:

Example 10. AB is a diameter. If ∠DAB = 300, ∠CBA = 100, then what is the degree measure of minor arc DC?

A. 900                     B. 600                            C. 1000                 D. None of these

Solution:

Join DB. We get

So, Angle ADB = 900 (Angle in a semicircle)

In ?ADB, sum of all the angles must be 1800

Angle CBD = 180 – 90 – 30 - 10 = 500

Thus, arc DC subtends angle of 50 degrees at the point B, so it will subtend an angle of 100 degrees at the centre.

Example 11. In the figure shown, SR is the diameter. If ∠QPR= 280, ∠QRT = 360, then what is the measure of ∠PRS?

1.  200                    B. 260                          C. 280                  D. 360

Solution: As SQ is the diameter, so angle SQR = 900.

∠RPQ = ∠QSR = 280 (Angles on the same arc QR)

Consider ?QSR, sum of all the angles must be 1800.

So, ∠PRS = 180 – 90 – 28 – 36 = 260

Property 3:

Example 12. P is a point outside a circle and is 13 cm away from its centre. A secant drawn from the point P intersects the circle at points A and B in such a way that PA = 9 cm and AB = 7cm. The radius of the circle is:

1. 5.5 cm                2. 5 cm                3. 4 cm                4. 4.5 cm

Solution:

Extend PO to intersect the circle at N as shown in the figure

PA * PB = PM * PN

9 * (9+7) = (13-r)*(13+r)

144 = 169 – r2

r2 = 25  or r = 5

Example 13: AB & CD are two perpendicular chords. AE = x, BE = 4x, CE = x + 2 and ED = 2x + 6, what is the area of the circle?

Solution

AE  BE = CE  ED

x (4x) = (x+2)(2x+6)

4x2 = 2x2 + 6x + 4x + 12

2x2 = 10x + 12

x2 – 5x – 6  = 0

x = 6

If we draw the perpendicular bisectors of AB & CD, then they will intersect at the center of the circle.

OA is the radius of the circle.

In AOX, applying Pythagoras theorem

AX2 + OX2 = AO2

152 + 52 = r2

250 = r2

Example 14: OR is a tangent to the circle of P at RS is the tangent at S. also O is the center of the circle. Given OP = 12, OQ = 6 what is the length of PR?

Solution

PR = RS [tangents from an external point are equal length]

OP2 = OQ * OS

122 = 6(6+2r)

24 = 6+ 2r  or  r = 9

Applying Pythagoras theorem in ?ORS

OR2 = OS2 + RS2

(12+x)2 = 242 + x2

144 + x2 + 24x = 576 + x2

x = 18

Another Method

2r = 18

R = 9

X = 2r

X = 18

PR = 18 units

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