HOTS Questions on Chapter 10 Class 10 Circles

HOTS Questions on Class 10 Maths Chapter 10 Circles help students understand tangents, chords, and angle properties in a simple way. They improve knowledge of theorems, circle geometry, and problem solving based on real life figures. This chapter builds a strong base for solving geometry problems and improving math skills. A downloadable PDF is also available for easy revision and practice.

HOTS Questions Based on Tangents to a Circle

Question 1: A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 13 cm. Find the length of PQ. Then find the angle that OQ makes with the tangent.

Answer:

Given:

Radius OP = 5 cm

OQ = 13 cm

∠OPQ = 90° (radius ⊥ tangent)

Using Pythagoras theorem in △OPQ:

OQ² = OP² + PQ²

13² = 5² + PQ²

169 = 25 + PQ²

PQ² = 144

PQ = 12 cm

For angle OQP:

tan(∠OQP) = OP/PQ = 5/12

∠OQP = tan⁻¹(5/12) ≈ 22.6°

Answer: PQ = 12 cm, angle ≈ 22.6°

Question 2: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. Find the radius of the circle. If another point R is located such that QR = 7 cm and R lies on the tangent line extended, what is the distance from R to the center?

Answer:

Part 1:

QT = 24 cm (tangent length)

OQ = 25 cm

∠OTQ = 90°

OT² = OQ² - QT²

OT² = 625 - 576 = 49

OT = 7 cm (radius)

Part 2:

R lies on tangent line extended, QR = 7 cm

OR² = OT² + TR²

T is point of tangency, TR = QT + QR? No:

If R is beyond Q on the tangent line:

QR = 7 cm, QT = 24 cm

RT = QT + QR = 31 cm (if R is beyond Q)

OR² = OT² + TR²

OR² = 49 + 961 = 1010

OR = √1010 ≈ 31.78 cm

Answer: Radius = 7 cm, OR ≈ 31.78 cm

Question 3: Prove that the tangents drawn at the ends of a diameter of a circle are parallel to each other.

Answer:

Given:

Circle with center O, diameter AB

PA is tangent at A

QB is tangent at B

To prove: PA || QB

Since OA is radius and PA is tangent at A:

OA ⊥ PA → ∠OAP = 90°

Since OB is radius and QB is tangent at B:

OB ⊥ QB → ∠OBQ = 90°

Now ∠OAP + ∠OBQ = 90° + 90° = 180°

These are co-interior angles formed by

transversal AB with lines PA and QB.

Since co-interior angles sum = 180°:

PA || QB Hence Proved.

Question 4: In the figure, XY and X'Y' are two parallel tangents to a circle with center O and AB is another tangent with point of contact C between XY and X'Y'. Prove that ∠AOB = 90°.

Answer:

Given:

XY tangent at P (top)

X'Y' tangent at Q (bottom)

AB tangent at C

OA bisects ∠POC and OB bisects ∠QOC.

In △OPA and △OCA:

OP = OC (radii)

AP = AC (tangents from A)

OA = OA (common)

By SSS: △OPA ≅ △OCA

So: ∠POA = ∠COA

OA bisects ∠POC

Similarly OB bisects ∠QOC.

∠POQ = 180° (P and Q are ends of diameter)

∠POC + ∠QOC = 180°

2∠AOC + 2∠BOC = 180°

∠AOC + ∠BOC = 90°

∠AOB = 90° Hence Proved

HOTS Questions Based on Tangent-Radius Theorem

Question 5: A student says "If the radius of a circle is doubled, the length of the tangent from the same external point also doubles." Is this correct? Justify with calculation.

Answer:

This statement is INCORRECT.

Let external point P be at distance d from center O.

Original: radius = r

Tangent length = √(d² - r²)

If radius is doubled: radius = 2r

New tangent = √(d² - 4r²)

Example: d = 10, r = 6

Original tangent = √(100 - 36) = √64 = 8

If radius doubled to 12:

New tangent = √(100 - 144) = √(-44)

This is not even real! The point would be

INSIDE the circle no tangent possible.

Even for valid case: d = 13, r = 5

Original tangent = √(169 - 25) = √144 = 12

Radius doubled to 10:

New tangent = √(169 - 100) = √69 ≈ 8.3

8.3 ≠ 24 (not doubled!)

The student is WRONG.

Question 6: Two concentric circles have radii 5 cm and 13 cm. Find the length of the chord of the larger circle that is tangent to the smaller circle.

Answer :

   (chord AB is tangent to small circle at M)

Since AB is tangent to small circle at M:

OM ⊥ AB → OM = 5 cm (radius of small circle)

OA = 13 cm (radius of large circle)

In right △OMA:

AM² = OA² - OM²

AM² = 169 - 25 = 144

AM = 12 cm

Since OM ⊥ AB, M is midpoint of AB:

AB = 2 × AM = 2 × 12 = 24 cm

Answer: Length of chord = 24 cm

Question 7: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.

Answer:

Given:

PA and PB are tangents from external point P

A and B are points of contact

O is center

To prove: ∠APB + ∠AOB = 180°

In quadrilateral OAPB:

∠OAP = 90° (radius ⊥ tangent at A)

∠OBP = 90° (radius ⊥ tangent at B)

Sum of angles in quadrilateral = 360°

∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

∠AOB + 90° + ∠APB + 90° = 360°

∠AOB + ∠APB = 360° - 180°

∠AOB + ∠APB = 180° Hence Proved.

HOTS Questions Based on Equal Tangents

Question 8: In the figure, a triangle ABC is circumscribed about a circle of radius r. If AB = 10 cm, BC = 8 cm, CA = 7 cm, find the length of the tangents from each vertex.

Answer:

Let tangent from A = x

Let tangent from B = y

Let tangent from C = z

Since tangents from same external point are equal:

From A: AF = AE = x

From B: BF = BD = y

From C: CD = CE = z

AB = x + y = 10  ...(1)

BC = y + z = 8   ...(2)

CA = z + x = 7   ...(3)

Adding all three equations:

2(x + y + z) = 25

x + y + z = 12.5  ...(4)

From (4) and (2): x = 12.5 - 8 = 4.5

From (4) and (3): y = 12.5 - 7 = 5.5

From (4) and (1): z = 12.5 - 10 = 2.5

Tangent from A = 4.5 cm

Tangent from B = 5.5 cm

Tangent from C = 2.5 cm

Question 9: From an external point P, two tangents PA and PB are drawn to a circle with center O. If ∠APB = 70°, find ∠OAB.

Answer:

Given: ∠APB = 70°

Find: ∠OAB

Step 1: In quadrilateral OAPB:

∠OAP = ∠OBP = 90° (radius ⊥ tangent)

∠AOB + ∠APB = 180° (proved earlier)

∠AOB = 180° - 70° = 110°

Step 2: In triangle OAB:

OA = OB = r (radii)

So △OAB is isosceles

∠OAB = ∠OBA

Step 3:

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB + 110° = 180°

2∠OAB = 70°

∠OAB = 35°

Answer: ∠OAB = 35°

Question 10: In a right triangle ABC, right-angled at B, a circle is inscribed. If AB = 6 cm, BC = 8 cm, find the radius of the inscribed circle.

Answer :

Step 1: Find hypotenuse AC

AC² = AB² + BC² = 36 + 64 = 100

AC = 10 cm

Step 2: Let radius = r

Let tangent from A = x

Let tangent from C = y

Let tangent from B = tangents = (6 − r) and (8 − r)

But from B: tangents = r (since ∠B = 90°)

So 6 − r + 8 − r = 10

14 − 2r = 10

2r = 4

r = 2 cm

Alternative formula:

r = (AB + BC − AC)/2

r = (6 + 8 − 10)/2

r = 4/2 = 2 cm

Answer: Radius = 2 cm

HOTS Questions Based on Circle Geometry

Question 11: Two circles of radii 5 cm and 3 cm touch externally. Find the length of the direct common tangent.

Answer:

For externally touching circles:

Distance between centers = r₁ + r₂ = 5 + 3 = 8 cm

Length of direct common tangent:

= √(d² - (r₁ - r₂)²)

= √(64 - (5-3)²)

= √(64 - 4)

= √60

= 2√15 cm

≈ 7.75 cm

Answer: Length = 2√15 cm ≈ 7.75 cm

Question 12: If PA and PB are tangents from point P to a circle with center O, and M is a point on arc AB (minor arc), find ∠AMB in terms of ∠APB.

Answer:

Given: ∠APB = θ

From earlier theorem:

∠AOB = 180° - θ

∠AMB is angle in the major arc

(M is on minor arc, so ∠AMB is in major segment)

By the inscribed angle theorem:

Angle in major segment = 180° - (angle in minor segment)

Reflex ∠AOB = 360° - (180° - θ) = 180° + θ

∠AMB = (1/2) × reflex ∠AOB

= (1/2)(180° + θ)

= 90° + θ/2

= 90° + ∠APB/2

Answer: ∠AMB = 90° + (∠APB/2)

Question 13: PQ is a chord of a circle and PT is tangent at P. If ∠TPQ = 60°, find ∠PRQ where R is a point on the major arc.

Answer:

By tangent-chord angle theorem (alternate segment):

∠TPQ = angle in alternate segment

∠PRQ = ∠TPQ = 60°

(The angle between tangent and chord equals

the angle subtended by chord in alternate segment)

Answer: ∠PRQ = 60°

Circle HOTS Formulas

• Tangent length from external point P: PT = √(OP² - r²)
• Equal tangents: PA = PB
• Tangent radius angle: ∠OTP = 90°
• Angle sum in quadrilateral OAPB: ∠AOB + ∠APB = 180°
• Tangent-secant: PA² = PB × PC

Inscribed circle radius (right triangle): r = (a + b - c)/2

where c is hypotenuse

For circumscribed quadrilateral: AB + CD = BC + DA

For tangent chord angle = angle in alternate segment

Download PDF - HOTS Questions on Class 10 Maths Chapter 10 Circles pdf

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