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# Geometry

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 45^{0} 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. 108^{0 } B. 120^{0} C. 90^{0} D. 150^{0}

**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, OB^{2} = OM^{2} + MB^{2}

5^{2 }= OM^{2} + 3^{2 }

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

πr^{2} = 64π or r = 8

In ?OAM

OA^{2} = OM^{2} + AM^{2}

8^{2} = 4^{2} + x^{2}

x^{2} = 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. 30^{0} 2. 45^{0} 3. 60^{0} 4. 90^{0}

**Solution: **

As OA = OB = AB = r (radius)

So, ?OAB is equilateral.

Thus, angle AOB = 60^{0}

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. 120^{0} 2. 30^{0} 3. 60^{0} 4. 90^{0}

**Solution: **

**Example 10.** AB is a diameter. If ∠DAB = 30^{0}, ∠CBA = 10^{0}, then what is the degree measure of minor arc DC?

A. 90^{0} B. 60^{0 } C. 100^{0 } D. None of these

**Solution: **

Join DB. We get

So, Angle ADB = 90^{0} (Angle in a semicircle)

In ?ADB, sum of all the angles must be 180^{0 }

Angle CBD = 180 – 90 – 30 - 10 = 50^{0}

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= 28^{0}, ∠QRT = 36^{0}, then what is the measure of ∠PRS?

1. 20^{0} B. 26^{0} C. 28^{0} D. 36^{0}

**Solution:** As SQ is the diameter, so angle SQR = 90^{0}.

∠RPQ = ∠QSR = 28^{0} (Angles on the same arc QR)

Consider ?QSR, sum of all the angles must be 180^{0}.

So, ∠PRS = 180 – 90 – 28 – 36 = 26^{0}

__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 – r^{2}

r^{2} = 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)

4x^{2} = 2x^{2} + 6x + 4x + 12

2x^{2} = 10x + 12

x^{2} – 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

AX^{2} + OX^{2} = AO^{2}

15^{2} + 5^{2} = r^{2}

250 = r^{2}

**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

12^{2} = 6(6+2r)

24 = 6+ 2r or r = 9

Applying Pythagoras theorem in ?ORS

OR^{2} = OS^{2 }+ RS^{2}

(12+x)^{2} = 24^{2} + x^{2 }

144 + x^{2} + 24x = 576 + x^{2}

x = 18

__Another Method __

2r = 18

R = 9

X = 2r

X = 18

PR = 18 units