QUESTIONS ON CIRCLE (PART-3)
QUESTIONS ON CIRCLE (PART-3)
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
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
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’)
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
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
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
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?
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’)
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’)
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
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’)
Two circles touch each other at a point A and PQ is a common tangent to both the circles; find ∠PAQ.
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’)
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
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’)
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
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’)
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 ?
OP = OA AP
=> OP = r – 2r/3
=> OP = r/3