CIRCLEGEOMETRYMATHS

QUESTIONS ON CIRCLE (PART-2)

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QUESTIONS ON CIRCLE (PART-2)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Geometry (Circle) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

QUERY 11

Radius of a circle is 10. Two chords of 16 cm and 17 cm cut each other perpendicularly; then distance between center of the circle and the intersecting point of two chords is ?

A) 6.5
B) 7.6
C) 8
D) 7.2

Vijay Bharath Reddy
The line joining the centre of the cirlce to the intersecting point of two chords is actually the diagonal of the rectangle formed by the altitudes drawn from centre to the two chords respectively.

We know that diagonal of a rectangle = √(length² + breadth²)
We see that the length and breadth of this rectangle actually are the altitudes drawn from centre to the two chords respectively.

So letting the altitude on the chord with 16 cm of length as H and the altitude on the chord with 17 cm of length as h we can write the above equation like this: diagonal = √(H² + h²)

But, H² = 10² – (16/2)² = 100 – 64 = 36
and h² = 10² – (17/2)² = 100 – 289/4 = 27.25

So according to the Pythagoras
the required difference = √(36 + 27.25) = √63.25 = 8 (appox)————option ‘C’


QUERY 12

AB is the diameter of a circle with centre O and P is a point on it. If ∠POA=120, then the value of ∠PBO=?

A) 60
B) 70
C) 55
D) 80

MAHA GUPTA
In the circle we get two triangles POA and POB;

In ΔPOA, ∠POA=120 (given)
Therefore in ΔPOB, ∠POB=180 – 120=60 (∠POB being the linear pair with ∠POA)

Therefore ∠PBO+∠OPB=180 – 60 = 120 (according to the angle sum property of a triangle)

Now ∠PBO = OPB (OB and PB being the radii of the same circle and hence equal; and both of these angles opposite to these sides)
Therefore ∠PBO and ∠OPB each equal to 60

So ∠PBO=60 (option ‘A’)


QUERY 13

The distance between the centers of two equal circles each of radius 3 cm is 10 cm. The length of a transverse tangent is?

A) 4 cm
B) 5 cm
C) 6 cm
D) 8 cm

MAHA GUPTA
Remember you can call TRANSVERSE TANGENT as TRANSVERSE COMMON TANGENT also.

Let A & B be the centers of the two circles respectively and PQ the transverse tangent intersecting the circle with center A at P & intersecting the circle with center B at Q and the line joining the centers at X.

Now X will be the bisector point of AB as a transverse tangent divides the line joining the centers internally in the ratio of their radii

Hence in triangle APX
AX = 10/2 = 5 cm
AP = 3 cm (radius)
∠APX = 90° (P being the point of contact of radius and tangent)

Hence by Pythagoras
5² – 3² = PX²
=> PX = 4
Hence PQ = 2*4 = 8 cm       (PQ =2*PX as it’s the transverse tangent intersecting the line joining the centers of two equal circles.) [option ‘D’]


QUERY 14

A circle and square have the equal perimeter, then

A) area of circle is greater
B) area of square is greater
C) both have equal areas
D) cannot be compared

SHIV KISHOR
2πr = 4a, where ‘r’ and ‘a’ are the radius of the circle and side of the square. (given)

Therefore area of the square a² = [2πr/4]²
=> (π²/4)r²
=> (π/4)(πr²)
Therefore a² < πr²

Thus area of circle is greater (option ‘A’)

TRICK
BHUSHAN AGGARWAL
The perimeters of a circle and a square are 2πr and 4a respectively, so we have to choose a number divisible by 22 and 4 both so that calculations are easy. Now let the perimeter is 44, therefore a = 11 and r = 7
Now area of the circle = (22/7)*7² = 154
And area of the square = 11² = 121
So area of circle is greater (option ‘A’)


QUERY 15

Two circles C1 and C2 of radii 2 cm and 3 cm respectively touch each other as shown in the figure. If AD and BD are tangents then the length of BD is?
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A) 15√10 cm
B) 1.5√10 cm
C) 10√10 cm
D) √10 cm

MAHA GUPTA
Let the point of intersection of the tangent AD to the circle C2 be T and the center of C2 be O and join O and T to form OT. Thus ΔOTA is a right angle triangle, where ∠OTA = 90°, AO = diameter of C1+radius of C2 = 4+3 = 7 and OT = radius of C2 = 3, and AO being the hypotenuse.

Therefore AT = √(7² – 3²) = √40 = 2√10 ……………by Pythagoras

Now, as BD is another tangent to C2, B being the point of tangent, and it’s joined with the center of the circle, so ΔABD too is a right triangle, where angle ABD = 90°, AB = sum of the diameters of the two circles given = 4+6 = 10 cm and AD being the hypotenuse.

Note that BT and TD are equal as they both are tangents to the same circle joined together (here joined at D)
Now let BD = x cm, therefore AD = (2√10 + x) cm

Hence by Pythagoras BD² = (2√10 +x)² – 10²

=> x² = 40 + x² + 4x√10 -100

=> 4x√10 – 60 = 0

=> x√10 = 15

=> x = 15/√10 = (15/10)√10 …………by rationalization of √10

=> x = 1.5√10 (option ‘B’)


QUERY 16

In a circle of radius 5 cm, AB and AC are two equal chords of lengths 6 cm each. Find the length of BC.

A) 10 cm
B) 8 cm
C) 9.6 cm
D) 10.8 cm

MAHA GUPTA
In a circle as O its center draw the figure according to the question. Join AO and OB. Obviously both are radii of the circle, means each is of length 5 cm.

Now you’ll see that ΔAOB is isosceles with AO and OB 5 cm each and AB 6 cm.

Now area of an isosceles triangle = (b/4)√(4a² – b²), where a is the length of one of the equal sides and b the third side.
Therefore area of ΔAOB = (6/4)√(4*5² – 6²)
= 12

We know that line going through the center and the intersecting point of the two equal chords is the perpendicular bisector of the line segment joining the two chords (here BC).

Now, Let this perpendicular bisecting point be M, so BM is the height of the ΔAOB corresponding to the base AO

Hence BM = (area/AO)*2 …………….(From area of the triangle = (base*height)/2)

=> BM = (12/5)*2 = 24/5 = 4.8 cm

But BM = 1/2 of BC…………….told above
Hence BC = 2*4.8 = 9.6 cm (option ‘C’)


QUERY 17

Let C be a point on a straight line AB. Circles are drawn with diameters AC and AB. Let P be any point on the circumference of the circle with diameter AB. If AP meets the other circle at Q, then?

A) QC || PB
B) QC is never parallel to PB
C) QC = 1/2 PB
D) QC || PB and QC = 1/2 PB

Sam Sampras
image

∠APB = ∠AQC = 90 (both are angles of semi- circles)
Now in ΔAPB and ΔAQC
∠A = ∠PAB = ∠QAC …….same angles
∠APB = ∠AQC …….proved above

Therefore both triangles are similar with ∠APB and ∠AQC being corresponding.
Hence QC ll PB

QC cannot necessarily be 1/2 PB as ‘C’ must be half of AB then, which cannot be proved

Hence QC ll PB (option ‘A’)


QUERY 18

N is the perpendicular from a point P of a circle with radius 7 cm on a diameter AB of the circle. If the length of the chord PB is 12 cm, the distance of the point N from the point B is?

A) 26/7
B) 72/7
C) 47/7
D) 86/7

SHIV KISHOR
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QUERY 19

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A) 4 cm
B) 3 cm
C) 2.5 cm
D) 5 cm

MAHA GUPTA
RULE: If two chords of a circle intersect inside, or outside the circle when produced the rectangle formed by the two segments of a chord is equal to the area to the rectangle formed by the two segments of the other chord.

Therefore AB*BP = CD*DP
=> CD = (5*3)/3 = 5 cm (option ‘D’)


QUERY 20

PQ is a chord of length 8 cm of a circle with center O and of radius 5 cm. The tangents at P and Q intersect at a point T. The length of TP is? 

A) 15/4 cm
B) 20/3 cm
C) 21/4 cm
D) 10/3 cm

SHIV KISHOR
In the following figure T is the point of intersection of the two tangents at P and Q. One of the tangents has not been shown as it was not needed in the solution.
1925193_550535715061461_68570796_n

Since the line joining the center of the circle to the intersecting points of the tangents at a chord cuts the chord on right angle; ∠OMP = 90°s and hence ΔOMP is a right angle triangle. Now,
1925193_550535715061461_68570796_n (1)

It can easily be made out that triangle OMP is similar to triangle OPT hence
1925193_550535715061461_68570796_n (2)

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QUESTIONS ON CIRCLE (PART-3)

Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
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