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All questions of Circles for Class 10 Exam

In the given figure, AR=5cm, BR=4cm and AC =11cm. What is the length of BC?
  • a)
    6 cm
  • b)
    10 cm
  • c)
    4 cm
  • d)
    8 cm
Correct answer is option 'B'. Can you explain this answer?

Vivek Rana answered
Since the sides of the triangles are tangent to the circle. BR = BP,AR = AQ and CP = CQ
So, BP = 4cm and AQ = 5
CQ = 11-5 = 6 = CP
BC = BP + CP = 4 + 6 = 10 cm
 

The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
  • a)
    √7 cm
  • b)
    2√7 cm
  • c)
    10 cm
  • d)
    5 cm
Correct answer is option 'B'. Can you explain this answer?

Pooja Shah answered
Let P be the external point and PA and PB be the tangents and OA and OB be the radii.
So OP is the hypotenuse=8cm 
Applying Pythagoras theorem,
H= P+ B2
64 = AP+ 36
AP = 

In fig, O is the centre of the circle, CA is tangent at A and CB is tangent at B drawn to the circle. If ∠ACB = 75°, then ∠AOB =
  • a)
    75°
  • b)
    85°
  • c)
    95°
  • d)
    105º
Correct answer is option 'D'. Can you explain this answer?

Solution:- The length of tangents drawn from an external point to the circle are equal.

In the figure, CA and CB are the tangents from external point of a circle and OA and OB are the two radius of a circle.

Draw a line OC, then you will get two triangles OAC and OBC.

In a ∆le AOC and ∆le BOC,

Angle OAC and OBC = 180 degree. Because, these are angles between the radii and tangents. So, there are two right angles.

The value for right angled triangle is 90 degree. Here, you can see both right angled triangle, so 90×2 = 180 degree.

Angle AOB = Angle OAC + Angle OBC - Angle ACB.

Given:- Angle ACB = 75 degree. , Angle OAC and OBC = 90 degree.

Angle AOB = 90 + 90 - 75.
Angle AOB = 180 -75.
Angle AOB =105 degree.

So, option d is correct friend.

A point P is 25 cm from the centre of a circle. The radius of the circle is 7 cm and length of the tangent drawn from P to the circle is x cm. The value of x =
  • a)
    20 cm
  • b)
    24 cm
  • c)
    18 cm
  • d)
    12 cm.
Correct answer is option 'B'. Can you explain this answer?

Given information
- A point P is 25 cm from the centre of a circle.
- The radius of the circle is 7 cm.
- Length of the tangent drawn from P to the circle is x cm.

To find
The value of x.

Solution
Let O be the centre of the circle and PT be the tangent drawn from P to the circle as shown below.

[INSERT IMAGE]

We can observe that OP is the hypotenuse of the right-angled triangle OPT. Therefore, using Pythagoras theorem, we can find the length of PT as follows.

OP² = OT² + PT²
(25)² = (7)² + PT²
625 = 49 + PT²
PT² = 625 - 49
PT² = 576
PT = √576
PT = 24 cm

Therefore, the value of x is 24 cm.

Answer: Option B.

If the diagonals of a cyclic quadrilateral are equal, then the quadrilateral is
  • a)
    rhombus
  • b)
    square
  • c)
    rectangle
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Raghav Bansal answered
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.

(Consider BD as a chord)
∠BCD + ∠BAD = 180 (Cyclic quadrilateral)

∠BCD = 180− 90 = 90
(Considering AC as a chord)

∠ADC + ∠ABC = 180 (Cyclic quadrilateral)

90+ ∠ABC = 180
∠ABC = 90

Each interior angle of a cyclic quadrilateral is of 90.Hence it is a rectangle.

 ABC is a right triangle right angled at B such that BC = 6 cm and AB = 8 cm. The radius of its incircle is
  • a)
    3 cm
  • b)
    8 cm
  • c)
    2 cm
  • d)
    5 cm
Correct answer is option 'C'. Can you explain this answer?

Coachify answered
By pythagoras theorem ,
H2 = P2 + B2
 =36 + 64=100
H=10
We have a square BPOQ so all sides are equal.
Since tangents from a point are equal in length
Let AP = AR = x
BP=BQ=y
RC = QC = z
x + y = 8
x = 8 - y
x + z = 10 ⇒ 8 - y + z = 10 ⇒ z = 2 + y
y + z = 6 ⇒ y + 2 + y = 6 ⇒ y = 2
x = 6, z= 4
Since BP = OQ = y = 2

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is a:
  • a)
    rectangle
  • b)
    square
  • c)
    parallelogram
  • d)
    cyclic quadrilateral
Correct answer is option 'D'. Can you explain this answer?

Cyclic quadrilaterals do not any definite rules unlike IIgram .The only rule that applies hear is that opp angles are supplementary.Hence it would'n matter if the angle bisectors form a quadrilateral    

AB is a chord of the circle and AOC is its diameter such that angle ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to
  • a)
     65°
  • b)
     60°
  • c)
    50°
  • d)
    40°
Correct answer is option 'C'. Can you explain this answer?

Understanding the Circle Geometry
In the given problem, we have a circle with a chord AB, and AOC is its diameter. The angle ACB is given as 50°. We need to find the measure of angle BAT, where AT is the tangent to the circle at point A.
Circle Properties and Tangents
- A tangent to a circle is perpendicular to the radius at the point of tangency.
- The angle between the tangent and the chord through the point of contact is equal to the angle in the alternate segment.
Applying the Theorem
- Since AT is the tangent at point A, and OA is the radius, we have angle OAT = 90°.
- According to the alternate segment theorem, angle BAT (the angle between the tangent AT and chord AB) is equal to angle ACB.
Calculating the Angles
- Given that angle ACB = 50°, we can write:
Angle BAT = Angle ACB
Therefore, Angle BAT = 50°
Conclusion
Thus, the measure of angle BAT is 50°, which corresponds to option 'c'. This is a fundamental property in circle geometry, illustrating the relationship between tangents and chords in a circle.

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is
  • a)
    12 cm
  • b)
    13 cm
  • c)
    8.5 cm
  • d)
    √119 cm
Correct answer is option 'D'. Can you explain this answer?

Kds Coaching answered
In the given problem, PQ is a tangent to the circle at point P, and OP is the radius to the point of tangency. By the property of tangents, OP is perpendicular to PQ, i.e., OP ⟂ PQ.
Given:
  • Radius, OP = 5 cm
  • OQ = 12 cm
Since OP ⟂ PQ, triangle OPQ is a right-angled triangle with the right angle at P.
Applying the Pythagorean theorem in triangle OPQ:
OQ² = OP² + PQ²
Substituting the known values:
12² = 5² + PQ²
144 = 25 + PQ²
PQ² = 144 - 25 = 119
Therefore, PQ = √119 cm.

Read the following text and answer the following question on the basis of the same:
A farmer has a field in the shape of triangle with AB = 13 cm, BC = 14 cm and AE = 7 cm. He wants to leave a space in the form of a circular field for growing wheat and the remaining for growing vegetables.
Q. The measure of AC is:
  • a)
    15 cm
  • b)
    17 cm
  • c)
    12 cm
  • d)
    13 cm
Correct answer is option 'A'. Can you explain this answer?

Radha Iyer answered
∴ AB = 13 cm (given)
BF = AB – AF
= (13 – 7) cm
AE = 7 cm (given)
Then, AF = AE (Tangents of the circle)
⇒ AF = 7 cm.
[Proved AF = 7 cm]
BF = 6 cm
BC = 14 cm (given)
and BD = 6 cm [Proved in Q. 3]
then CD = BC – BD
= (14 – 6) cm = 8 cm
but CD = CE (Tangents of the circle)
∴ CE = 8 cm
So, CD + CE = (8 + 8) cm = 16 cm.
CE = 8 cm
and AE = 7 cm (given)
then, AC = AE + EC
= (7 + 8) cm
= 15 cm.

Both Assertion and Reason are true but Reason is not correct explanation of Assertion
In the following figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm.
Q. Find the value of BP:
  • a)
    10 cm
  • b)
    11 cm
  • c)
    12 cm
  • d)
    13 cm
Correct answer is option 'B'. Can you explain this answer?

Ananya Das answered
AB = 29 cm (given) but AQ = AR(Length of tangents are equal) then AQ = 18 cm
Here, AD = 23 cm (given)
and
Here, DS = 5 cm
but DR = DS (Length of tangents are equal)
then DR = 5 cm.
DR = 5 cm (Proved)
then AR = AD – DR cm
= (23 – 5) cm = 18 cm. [Proved]
∴ BQ = AB – AQ
= (29 – 18) cm = 11 cm.
BQ = 11 cm [Proved]
and BP = BQ (Length of tangents are equal)
∴ BP = 11 cm.

In the figure, PQ is a tangent at a point R of the circle with centre O. If ∠TRQ = 30°.The measure of ∠PRS is
  • a)
    45°
  • b)
    120°
  • c)
    90°
  • d)
    60°
Correct answer is option 'D'. Can you explain this answer?

Arwin answered
TRQ=30,
TSR=TRQ=30( alternate segment theorem)-1
TRS=90( angle in semicircle)--2,
from 1&2 and by angle sum property of a triangle,
RTS=60,
since RTS=PRS ( by alternate segment theory)
PRS=60

In fig, O is the centre of the circle. PQ is tangent to the circle and secant PAB passes through the centre O. If PQ = 5 cm and PA = 1 cm, then the radius of the circle is
  • a)
    8 cm
  • b)
    12 cm
  • c)
    10 cm
  • d)
    6 cm
Correct answer is option 'B'. Can you explain this answer?

Kds Coaching answered
AOB is the diameter of the circle since POB passes through O.
Now PA is the segment of the secant POB outside the circle.
So, by the tangent–secant rule, the square of the tangent = secant segment × segment of the secant outside the circle when a secant and tangent to a circle are drawn from a point outside the circle.
PB × PA = PQ × PA ⇒ PB = (PQ²) / PA
PB = (25²) / 1 = 625 / 1 = 25 cm
The diameter AOB = PB - PA = (25 - 1) cm = 24 cm
The radius = x = AOB / 2 = 24 cm / 2 = 12 cm

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