Class 10 Maths Chapter 8 Circles Most Important Questions By PW

Physics Wallah provides a comprehensive set of Most Important Questions for Class 10 Maths Chapter 8 Circles, covering all key topics in the syllabus. These include tangents to a circle, the number of tangents from a point, lengths of tangents, and important theorem-based problems. You will also work on proofs, construction-based reasoning, and application-oriented questions commonly asked in exams. With PW’s targeted practice, you can improve conceptual clarity, reduce calculation errors, and develop a strong approach to solving a range of problems with confidence and accuracy for your upcoming board exams.

CBSE Class 10 Sample Paper

Class 10 Circles Most Important Questions by PW

Q.1: ABC is a right angled triangle, right angled at B such that BC = 6 cm, AB = 8 cm and AC = 10 cm. A circle with centre O is inscribed in ΔABC. The radius of the circle is:

Options

(a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm

Solution:

An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R. OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA. OPBQ is a square. ( ∵ ∠B - 90 ^o ) Let r be the radius of the circle PB = BQ = r AR = AP = 8 - r, CQ = CR = 6 - r AC = AR + CR ⇒ 10 = 8 - r + 6 - r ⇒ 10 = 14 - 2r ⇒ 2r = 14 - 10 = 4 ⇒ r = 2

Q.2: What should be the angle between the two tangents which are drawn at the end of two radii and are inclined at an angle of 45 degrees?

Solution:

The angle between them shall be 135 degrees.

Q.3: In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is

(A) 15 cm (B) 10 cm (C) 9 cm (D) 7.5 cm

Solution:

AQ, AP and BC are tangents to the circle with centre O and AB = 5cm , AC =6cm and BC = 4cm. Let BS = x then CS = 4 – x As, we know that tangents from an exterior point to a circle are equal in length ⇒AP=AQ ⇒ AB + BP = AC + CQ ........ (1) BP = BS and CQ = CS .........(2) from (1) and (2) we get AB + BS = AC + CS 5+ x = 6 + (4– x) ⇒ x + x = 6+4-5 ⇒ 2x = 5 ⇒ x = 5/2=2.5 Hence, AP = AB + BP = AB + BS = 5 cm + 2.5 cm = 7.5cm

Q.4: Shipra prepared a project for rain water harvesting. diagrammatic representation of the project is given in the figure. PQ and PR are the pipes touching the circular pit. Length of these pipes is 5 m each. What is the perimeter of ΔPMN?

Solution:

Here, MT = MQ [Tangents from point M] ...(i) And NT = NR [Tangents from point N] ...(ii) Now, PQ + PR = PM + MQ + PN + NR = PM + MT + PN + NT [Using eq. (i) and (ii)] = PM + PN + (MT + NT) = PM + PN + MN = Perimeter of DPMN ∴ Perimeter of △PMN = 5 + 5 = 10 cm.

Q.5: In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.

Solution:

∠ACB = 90° …[Angle in the semi-circle In ∆ABC, ∠CAB + ∠ACB + ∠CBA = 180° 30 + 90° + ∠CBA = 180° ∠CBA = 180° – 30° – 90° = 60° ∠PCA = ∠CBA …[Angle in the alternate segment ∴ ∠PCA = 60°

Q.6: I n the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then find the length of each tangent.

 

Solution:

Construction: Join AC and BC. Proof: ∠1 = ∠2 = 90° ….[Tangent is I to the radius (through the point of contact ∴ APBC is a square. Length of each tangent = AP = PB = 4 cm = AC = radius = 4 cm

Q.7: In the given figure, PQ and PR are two tangents to a circle with Centre O. If ∠QPR = 46°, then calculate ∠QOR.

Solution:

∠OQP = 900 ∠ORP = 90° ∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° …[Angle sum property of a quad. 90° + 46° + 90° + ∠QOR = 360° ∠QOR = 360° – 90° – 46° – 90° = 134°

 

Q.8: From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB.

 

Solution:

PA = PB …[∵ Tangents drawn from external point are equal∠PBA = ∠PAB = 50° …[Angles equal to opposite sides In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆ 50° + 50° + ∠APB = 180° ∠APB = 180° – 50° – 50° = 80° In cyclic quadrilateral OAPB ∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180° ∠AOB + 80o = 180° ∠AOB = 180° – 80° = 100°

 

Q.9: If two tangents inclined at an angle 60 ∘ are drawn to a circle of radius 3 c m, then the length of each tangent is

 

(A) 3 2 √ 3 c m (B) 6 c m (C) 3 c m (D) 3 √ 3 c m

 

Solution:

 

Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60 ∘

Radius of the circle = 3 c m Join OA and OP Also, OP is a bisector line of ∠ APC ∴ ∠ A P O = ∠ C P O = 30 ∘ O A ⊥ A P Also, tangents at any point of a circle is perpendicular to the radius through the point of contact. In right angled Δ O A P , we have tan 30 ∘ = O A A P = 3 A P ⇒ 1 √ 3 = 3 A P ⇒ A P = 3 √ 3 c m A P = C P = 3 √ 3 c m [Tangents drawn from an external point are equal] 

Hence, the length of each tangent is 3 √ 3 c m .

Q.10: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution:

Given: A circle with center O

with tangent X Y at point of contact P .

 

To Prove: O P ⊥ X Y

 

Proof: Let Q be a point on X Y connect O Q

 

Suppose it touches the circle at R

 

Hence,

 

O Q > O R

 

O Q > O P ( ∵ O P = O R ) (radius)

 

Same will be the case with all other points on the circle

 

Hence,

 

We get O P is the smallest line that connects X Y .

 

Q.11. Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r.

 

Solution:

 

Given: OC = 7 cm, AB = 48 cm To find: r = ? ∠OCA = 90° ..[Tangent is ⊥ to the radius through the point of contact ∴ OC ⊥ AB AC = 1 2 (AB) … [⊥ from the centre bisects the chord ⇒ AC = 1 2 (48) = 24 cm In rt. ∆OCA, OA ² = OC ² + AC ² … [Pythagoras’ theorem r ² = (7) ² + (24) ² = 49 + 576 = 625 ∴ r= 625 −−−√ = 25 cm

 

Q.12: Prove that the parallelogram circumscribing a circle is a rhombus.

 

Solution:

 

ABCD is a ॥ ^gm . To prove. ABCD is a rhombus. Proof. In ॥ ^gm , opposite sides are equal

AB = CD and AD = BC ..(i) AP = AS …[Tangents drawn from an external point are equal in length PB = BQ CR = CO DR = DS By adding these tangents, (AP + PB) + (CR + DR) = AS + BQ + CQ + DS AB + CD = (AS + DS) + (BQ + CQ) AB + CD = AD + BC AB + AB = BC + BC … [From (i) 2AB = 2 BC AB = BC …(ii) From (i) and (ii), AB = BC = CD = DA ∴ ॥ ^gm ABCD is a rhombus.

 

Q. 13: In the given figure, an isosceles ∆ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

Solution:

 

The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively.AB = AC To prove: BD = CD Proof: Since the lengths of tangents drawn from an external point to a circle are equal ∴ AF = AE … (i) BF = BD …(ii) CD = CE …(iii) Adding (i), (ii) and (iii), we get AF + BF + CD = AE + BD + CE ⇒ AB + CD = AC + BD But AB = AC … [Given ∴ CD = BD

 

Q.14: In the given figure, a circle inscribed in ∆ABC touches its sides AB, BC and AC at points D, E & F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.

Solution:

Let AD = AF = x BD = BE = y …[Two tangents drawn from and an external point are equal CE = CF = z

 

AB = 12 cm …[Given ∴ x + y = 12 cm …(i) Similarly, y + z = 8 cm …(ii) and x + z = 10 cm …(iii) By adding (i), (ii) & (iii) 2(x + y + z) = 30 x + y + z = 15 …[∵ x + y = 12 z = 15 – 12 = 3 Putting the value of z in (ii) & (iii), y + 3 = 8 y = 8 – 3 = 5 x + 3 = 10 x = 10 – 3 = 7 ∴ AD = 7 cm, BE = 5 cm, CF = 3 cm

 

Q: 15. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.

 

Solution:

 

The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively.

AB = AC To prove: BD = CD Proof: AF = AE ..(i) BF = BD …(ii) CD = CE …(iii) Adding (i), (ii) and (iii), we get AF + BF + CD = AE + BD + CE ⇒ AB + CD = AC + BD But AB = AC …[Given ∴ CD = BD]