CIRCLEGEOMETRYMATHS

QUESTIONS ON CIRCLE (PART-3)

QUESTIONS ON CIRCLE (PART-3)

QUERY 21

Two circles of radii 8 and 2 cm touch each other externally at point A. PQ is direct common tangent. Length of QP is?

A) 8 cm
B) 7.5 cm
C) 7 cm
D) 9 cm

MAHA GUPTA
10255426_226258567569961_7904956742858401990_n

Draw QR || ST
So, QR = ST = 10 cm

Now in right triangle QPR
PR = PS – RS = 8 – 2 = 6 cm
Diagonal QR = 10 cm

Therefore by Pythagoras, QP = √(10² – 6²) = 8 cm (option ‘A’)


QUERY 22

One chord of a circle is known to be 10.1 cm. The radius of this circle must be?

A) 5 cm
B) greater than 5 cm
C) greater than or equal to 5 cm
D) less than 5 cm

MAHA GUPTA
The longest chord of the circle is its diameter
Means the diameter cannot be bigger than 10.1 cm, Now letting the diameter of the circle be 10.1 cm
Therefore the radius = 10.1/2 = 5.5 cm

Hence option ‘B’ is correct


QUERY 23

P is a point outside a circle and is 13 cm away from its center. 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 = 7 cm. The radius of the circle is?

A) 5 cm
B) 4 cm
C) 4.5 cm
D) 5.5 cm

SHIV KISHOR
10001548_563991707049195_1454591722979390004_n

(Option ‘A’)

QUERY 24

In a circle, AB is diameter, CD is chord of length equal to radius of the circle. AC and BD when produced intersect at point E. Find ∠AEB?

A) 60°
B) 65°
C) 70°
D) 55°

MAHA GUPTA
Image0017

Triangle ODC is equilateral. (OC and OD being the radii of the circle and CD being given equal to the radius)

Therefore, ∠COD = 60°

Now, ∠CBD =½∠ COD (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Here arc CD subtending ∠CBD at the center and ∠COD on any other part of the circle)

=> ∠CBD = 30°

Now ∠ACB = 90° (angle in the semi-circle)

So, ∠BCE = 180° – ∠ACB = 90°

Therefore ∠CEB = 90° – 30° = 60°, i.e. ∠AEB = 60° (option ‘A’)


QUERY 25

10259997_1439070756341580_1579791458848474113_n

A) 80°
B) 85°
C) 70°
D) 75°

MAHA GUPTA
The arch RS is subtending the ∠ROS = 40 at the centre and the ∠RQS an other angle on the remaining part of the circle; therefore ∠RQS = ½∠ ROS (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.)
=> ∠RQS = 40/2 = 20°

Now ∠PRQ = 90° (angle in the semi-circle)

So, ∠QRT = 180° – ∠PRQ = 90°

Therefore ∠RTQ = 90° – 20° = 70°, i.e. ∠RTS = 70° (option ‘C’)


QUERY 26

O is centre of a circle. Two cords AC and BD intersect at P. If ∠AOB =15 and ∠APB=30, find tan² ∠APB + Cot² ∠COD

A) 4/3
B) 1/3
C) 2/3
D) 1

MEGHA DIXIT
10247286_1414796295458066_7820068920126733130_n (1)    

As ∠AOB = 15, hence ∠ADB = 15/2

∠APD = 150

Now in ΔAPD, ∠P = 150, ∠D = 15/2
Therefore ∠A = 45/2 and ∠DOC = 45

Then tan² 30 + cot² 45 = 1/3 + 1 = 4/3 (option ‘A’)

QUERY 27

Two circles touch each other at a point A and PQ is a common tangent to both the circles; find ∠PAQ.

A) 60°
B) 75°
C) 80°
D) 90°

SHIV KISHOR
If two circles, either equal or not, when touch at a point, say A, their common tangent, say PQ always makes angle PAQ = 90 (option ‘D’)


QUERY 28

Radii of two circles are 7 cm and 3 cm respectively. If one of those circles is completely inside the other circle, what will be the distance between the centers?

A) 4 cm more
B) more than 5 cm
C) less than 4 cm
D) none of the above

MAHA GUPTA
Distance between the centers of two circles is the difference of their radii if one of them is touching the other internally.

But here the circles are not touching, rather one of them lies completely inside of the other, therefore the distance between the centers will be less than the difference of their radii, means less than (7 – 3) i.e. 4 (option ‘C’)


QUERY 29

ABC is an isosceles triangle with AB = AC. A circle through B touching AC at the middle point intersects AB at P. Then AP : AB is?

A) 4 : 1
B) 2 : 3
C) 3 : 5
D) 1 : 4

SHIV KISHOR
10341588_594117757369923_4351242708453599008_n-1

AM is a tangent to the circle and APB is a secant to it. Thus by tangent theorem AM² = AP × AB

But AM = AC/2 = AB/2

=> AB²/4 = AP × AB

=> AP/AB = 1/4 = 1 : 4 (option ‘D’)

QUERY 30

AB = 2r is the diameter of a circle. If a chord CD intersect AB at right angle at point P in the ratio 1 : 2, then CD is ?

A) 4√2r/3
B) 2√2r/3
C) 3√2r/4
D) √2r/3

SHIV KISHOR
10306555_597818910333141_657988054765243785_n

AB = 2r
The chord CD intersect AB (2r) at point P in the ratio 1 : 2
Means AP : PB = 1 : 2
=> AP = 2r/3

OP = OA AP
=> OP = r – 2r/3
=> OP = r/3

Now in right triangle OPC
10306555_597818910333141_657988054765243785_n (2)

(Option ‘A’)

Previous post

QUESTIONS ON CIRCLE (PART-2)

Next post

QUESTIONS ON CIRCLE (PART-4)

Maha Gupta

Maha Gupta

Founder of SSCEXAMFORUM.COM and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC.

No Comment

Leave a reply