Important Questions Class 10 Maths Chapter 10: Circles With Solutions

A tangent to a circle is a line that touches the circle at exactly one point, called the point of contact. A secant cuts the circle at two distinct points, and a non-intersecting line has no common point with the circle.

Two theorems control every scoring pattern in Circles. CBSE 2026 tests Important Questions Class 10 Maths Chapter 10 through tangent properties, radius-tangent perpendicularity, equal tangents, angle calculations, and length-based problems. The NCERT Reprint 2026-27 syllabus includes 17 NCERT questions from Chapter 10. Students who know Theorem 10.1 and Theorem 10.2 can solve MCQs, short answers, proof questions, and tangent from external point class 10 questions with clear steps.

Key Takeaways

• Circles: Chapter 10 focuses on tangents, radius-tangent relation, equal tangents, and theorem-based proofs.
• Main Theorem: The radius drawn to the point of contact is always perpendicular to the tangent.
• Exam Trap: The angle between two tangents and the angle at the centre are supplementary.
• Tangent Count: A point inside a circle has 0 tangents, a point on the circle has 1, and an external point has 2.

Important Questions Class 10 Maths Chapter 10 Structure 2026

Important Questions Class 10 Maths Chapter 10: Theorems That Drive Every Question

These theorem on circles class 10 important questions follow the CBSE 2026 syllabus. Students must know the statement and use of each theorem.

Q1. What Is Theorem 10.1 In Class 10 Circles?

Theorem 10.1 states that the tangent at any point of a circle is perpendicular to the radius through the point of contact. If OP is the radius and PT is the tangent, then OP ⟂ PT.

1. Given Data:
   O is the centre.
   P is the point of contact.
   PT is the tangent.
2. Formula Used:
   OP ⟂ PT
3. Final Result:
   Angle OPT = 90°

Q2. What Is Theorem 10.2 In Class 10 Circles?

Theorem 10.2 states that tangents drawn from an external point to a circle are equal in length. If PA and PB are tangents from P, then PA = PB.

1. Given Data:
   P is an external point.
   PA and PB are tangents.
2. Formula Used:
   PA = PB
3. Final Result:
   Tangents from an external point are equal

Q3. How Many Tangents Can Be Drawn From A Point To A Circle?

The number of tangents depends on the point’s position. A point inside has 0 tangents, a point on the circle has 1 tangent, and a point outside has 2 tangents.

1. Point Inside Circle:
   0 tangents
2. Point On Circle:
   1 tangent
3. Point Outside Circle:
   2 tangents
4. Final Result:
   Number of tangents depends on the point’s position

Circles Class 10 MCQ With Answers

These circles class 10 MCQ questions test tangent properties, radius use, and Pythagoras theorem. Each question follows the CBSE 2026 board exam pattern.

Q1. From Point Q, The Tangent Length Is 24 cm And Distance From Centre Is 25 cm. Find The Radius.

The radius is 7 cm. The radius and tangent form a right angle at the point of contact.

1. Given Data:
   OQ = 25 cm
   PQ = 24 cm
   OP = r
2. Formula Used:
   OQ^2 = OP^2 + PQ^2
3. Calculation:
   25^2 = r^2 + 24^2
   625 = r^2 + 576
   r^2 = 49
4. Final Result:
   Answer: (A) 7 cm

Q2. TP And TQ Are Tangents With Centre O. If Angle POQ = 110°, Find Angle PTQ.

Angle PTQ is 70°. The radius is perpendicular to the tangent at both contact points.

1. Given Data:
   Angle POQ = 110°
   Angle OPT = 90°
   Angle OQT = 90°
2. Formula Used:
   Sum of angles in quadrilateral POQT = 360°
3. Calculation:
   Angle PTQ = 360° - 90° - 90° - 110°
   Angle PTQ = 70°
4. Final Result:
   Answer: (B) 70°

Q3. Tangents PA And PB From Point P Make Angle APB = 80°. Find Angle POA.

Angle POA is 50°. The line joining centre and external point bisects the angle between equal tangents.

1. Given Data:
   Angle APB = 80°
   Angle OAP = 90°
   Angle OBP = 90°
2. Formula Used:
   Angle AOB + Angle APB = 180°
3. Calculation:
   Angle AOB = 180° - 80°
   Angle AOB = 100°
   Angle POA = 100°/2
4. Final Result:
   Answer: (A) 50°

Q4. Radii Of Two Concentric Circles Are 4 cm And 5 cm. Find The Chord Of The Larger Circle Tangent To The Smaller Circle.

The chord length is 6 cm. The perpendicular from the centre bisects the chord.

1. Given Data:
   Larger radius = 5 cm
   Smaller radius = 4 cm
2. Formula Used:
   Half chord = sqrt(5^2 - 4^2)
3. Calculation:
   Half chord = sqrt(25 - 16)
   Half chord = sqrt(9) = 3 cm
   Chord = 2 × 3
4. Final Result:
   Answer: (B) 6 cm

Q5. From P, Distance From Centre O Is 13 cm And Radius Is 5 cm. Find The Tangent Length.

The tangent length is 12 cm. Radius and tangent form a right triangle with OP as hypotenuse.

1. Given Data:
   OP = 13 cm
   OQ = 5 cm
   PQ = tangent length
2. Formula Used:
   PQ = sqrt(OP^2 - OQ^2)
3. Calculation:
   PQ = sqrt(13^2 - 5^2)
   PQ = sqrt(169 - 25)
   PQ = sqrt(144)
4. Final Result:
   Answer: 12 cm

Q6. Tangent PQ At P Meets A Line Through O At Q. If OP = 5 cm And OQ = 12 cm, Find PQ.

The tangent length PQ is sqrt(119) cm. Use Pythagoras theorem in right triangle OPQ.

1. Given Data:
   OP = 5 cm
   OQ = 12 cm
2. Formula Used:
   PQ^2 = OQ^2 - OP^2
3. Calculation:
   PQ^2 = 12^2 - 5^2
   PQ^2 = 144 - 25
   PQ^2 = 119
4. Final Result:
   Answer: sqrt(119) cm

Q7. How Many Tangents Can Be Drawn From A Point Inside A Circle?

A point inside a circle has 0 tangents. Every line through that point cuts the circle at two points.

1. Given Data:
   Point lies inside the circle.
2. Rule Used:
   Tangent touches the circle at exactly one point.
3. Final Result:
   Answer: (A) 0 tangents

Q8. What Is The Maximum Number Of Parallel Tangents A Circle Can Have?

A circle can have 2 parallel tangents in one direction. They touch the circle at opposite ends of a diameter.

1. Given Data:
   One circle
2. Rule Used:
   Parallel tangents touch opposite points.
3. Final Result:
   Answer: (B) 2

Q9. If PA And PB Are Tangents From P And PA = 9 cm, Find PB.

PB is 9 cm. Tangents from the same external point have equal lengths.

1. Given Data:
   PA = 9 cm
   PA and PB are tangents from P
2. Formula Used:
   PA = PB
3. Final Result:
   PB = 9 cm

Q10. If A Point Lies On A Circle, How Many Tangents Can Pass Through It?

Exactly 1 tangent can pass through a point on a circle. The tangent touches the circle at that point only.

1. Given Data:
   Point lies on the circle.
2. Rule Used:
   One tangent exists at each point of a circle.
3. Final Result:
   Answer: 1 tangent

Tangent To A Circle Class 10 Important Questions With Short Answers

These tangent to a circle class 10 important questions follow 2-mark and 3-mark formats. Most answers use Theorem 10.1 with Pythagoras theorem.

Q1. Two Concentric Circles Have Radii 5 cm And 3 cm. Find The Chord Of The Larger Circle Touching The Smaller Circle.

The chord length is 8 cm. The radius to the point of contact stands perpendicular to the tangent chord.

1. Given Data:
   OA = 5 cm
   OP = 3 cm
   AB is tangent to the smaller circle at P
2. Formula Used:
   OA^2 = OP^2 + AP^2
3. Calculation:
   5^2 = 3^2 + AP^2
   25 = 9 + AP^2
   AP^2 = 16
   AP = 4 cm
4. Final Result:
   AB = 2 × AP = 8 cm

Q2. From Point A At Distance 5 cm From The Centre, The Tangent Length Is 4 cm. Find The Radius.

The radius is 3 cm. Radius and tangent form a right angle at the point of contact.

1. Given Data:
   OA = 5 cm
   AB = 4 cm
   OB = r
2. Formula Used:
   OA^2 = OB^2 + AB^2
3. Calculation:
   5^2 = r^2 + 4^2
   25 = r^2 + 16
   r^2 = 9
4. Final Result:
   Radius = 3 cm

Q3. If The Angle Between Two Tangents From P Is 90°, Find OP In Terms Of Radius a.

OP equals a√2. The centre line bisects the angle between two equal tangents.

1. Given Data:
   Angle between tangents = 90°
   Radius = a
2. Formula Used:
   sin 45° = a/OP
3. Calculation:
   1/√2 = a/OP
   OP = a√2
4. Final Result:
   OP = a√2

Q4. From Point Q, Tangents QA And QB Are Drawn. If Angle AQB = 60°, Find Angle OAB.

Angle OAB is 30°. Equal tangents form an isosceles triangle.

1. Given Data:
   Angle AQB = 60°
   QA = QB
2. Formula Used:
   OA ⟂ QA
3. Calculation:
   Triangle QAB is equilateral.
   Angle QAB = 60°
   Angle OAQ = 90°
   Angle OAB = 90° - 60°
4. Final Result:
   Angle OAB = 30°

Q5. If OP = 10 cm And Radius OT = 6 cm, Find The Tangent PT.

The tangent length PT is 8 cm. Apply Pythagoras theorem in right triangle OPT.

1. Given Data:
   OP = 10 cm
   OT = 6 cm
2. Formula Used:
   PT = sqrt(OP^2 - OT^2)
3. Calculation:
   PT = sqrt(10^2 - 6^2)
   PT = sqrt(100 - 36)
   PT = sqrt(64)
4. Final Result:
   PT = 8 cm

Circles Class 10 Proof Questions With Solutions

These circles class 10 proof questions test theorem statements, construction, and reasoning. Write every proof in short logical steps.

Q1. Prove That Tangents Drawn At The Ends Of A Diameter Are Parallel.

The tangents are parallel because both stand perpendicular to the same straight line. The diameter acts as the common transversal.

1. Given Data:
   AB is a diameter.
   PQ is tangent at A.
   RS is tangent at B.
2. To Prove:
   PQ ∥ RS
3. Proof:
   OA ⟂ PQ because radius is perpendicular to tangent.
   OB ⟂ RS because radius is perpendicular to tangent.
   AB is the same straight line through O.
   Both tangents are perpendicular to AB.
4. Final Result:
   PQ ∥ RS

Q2. Prove That Tangents From An External Point To A Circle Are Equal.

The tangents are equal in length because two right triangles become congruent by RHS.

1. Given Data:
   P is an external point.
   PQ and PR are tangents.
   O is the centre.
2. To Prove:
   PQ = PR
3. Construction:
   Join OP, OQ, and OR.
4. Proof:
   Angle OQP = 90°
   Angle ORP = 90°
   OQ = OR because both are radii.
   OP = OP because it is common.
   Triangle OQP ≅ Triangle ORP by RHS.
5. Final Result:
   PQ = PR by CPCT

Q3. Prove That AB + CD = AD + BC In A Quadrilateral Circumscribing A Circle.

The result follows from equal tangents drawn from the same external point. Each vertex gives one pair of equal tangent lengths.

1. Given Data:
   ABCD circumscribes a circle.
   The circle touches AB, BC, CD, and AD.
2. Formula Used:
   Tangents from the same external point are equal.
3. Proof:
   From A: AP = AS
   From B: BP = BQ
   From C: CQ = CR
   From D: DR = DS
4. Adding:
   AP + BP + CR + DR = AS + DS + BQ + CQ
   AB + CD = AD + BC
5. Final Result:
   AB + CD = AD + BC

Q4. Prove That The Angle Between Two Tangents Is Supplementary To The Angle At The Centre.

The two angles are supplementary because the radius-tangent angles are both 90°.

1. Given Data:
   PA and PB are tangents.
   A and B are points of contact.
   O is the centre.
2. To Prove:
   Angle APB + Angle AOB = 180°
3. Proof:
   Angle OAP = 90°
   Angle OBP = 90°
   In quadrilateral OAPB:
   Angle OAP + Angle APB + Angle OBP + Angle AOB = 360°
   90° + Angle APB + 90° + Angle AOB = 360°
4. Final Result:
   Angle APB + Angle AOB = 180°

Q5. Prove That A Parallelogram Circumscribing A Circle Is A Rhombus.

The parallelogram becomes a rhombus because adjacent sides become equal through tangent lengths.

1. Given Data:
   ABCD is a parallelogram.
   It circumscribes a circle.
2. To Prove:
   ABCD is a rhombus.
3. Proof:
   Tangents from the same external point are equal.
   So, AB + CD = BC + AD.
   In a parallelogram, AB = CD and BC = AD.
   Therefore, 2AB = 2BC.
   AB = BC.
4. Final Result:
   ABCD is a rhombus

Length Of Tangent Class 10 Questions With Solutions

These length of tangent class 10 questions combine Theorem 10.1 and Pythagoras theorem. Each solution uses a right triangle.

Q1. From Point P, A Tangent PA Is Drawn To A Circle With Centre O. If OP = 13 cm And OA = 5 cm, Find PA.

The tangent PA is 12 cm. Radius OA is perpendicular to tangent PA.

1. Given Data:
   OP = 13 cm
   OA = 5 cm
2. Formula Used:
   PA = sqrt(OP^2 - OA^2)
3. Calculation:
   PA = sqrt(13^2 - 5^2)
   PA = sqrt(169 - 25)
   PA = sqrt(144)
4. Final Result:
   PA = 12 cm

Q2. If Tangents PA And PB From P Are 10 cm Each, Find The Perimeter Of Triangle PAB When AB = 12 cm.

The perimeter of triangle PAB is 32 cm. Equal tangents give PA = PB.

1. Given Data:
   PA = 10 cm
   PB = 10 cm
   AB = 12 cm
2. Formula Used:
   Perimeter = PA + PB + AB
3. Calculation:
   Perimeter = 10 + 10 + 12
4. Final Result:
   Perimeter = 32 cm

Q3. From P, Tangents PA And PB Touch A Circle At A And B. If Angle APB = 60° And PA = 8 cm, Find PB.

PB is 8 cm. Tangents from the same external point are equal.

1. Given Data:
   PA = 8 cm
   PA and PB are tangents from P
2. Formula Used:
   PA = PB
3. Final Result:
   PB = 8 cm

Q4. If OP = 25 cm And Radius OA = 7 cm, Find Tangent PA.

The tangent PA is 24 cm. Use Pythagoras theorem in right triangle OAP.

1. Given Data:
   OP = 25 cm
   OA = 7 cm
2. Formula Used:
   PA = sqrt(OP^2 - OA^2)
3. Calculation:
   PA = sqrt(25^2 - 7^2)
   PA = sqrt(625 - 49)
   PA = sqrt(576)
4. Final Result:
   PA = 24 cm

Q5. If A Chord Of A Larger Circle Touches A Smaller Concentric Circle, How Do You Find Its Length?

Find half the chord using sqrt(R^2 - r^2) and double it. Here, R is the larger radius and r is the smaller radius.

1. Given Data:
   Larger radius = R
   Smaller radius = r
2. Formula Used:
   Half chord = sqrt(R^2 - r^2)
3. Calculation:
   Chord length = 2 × sqrt(R^2 - r^2)
4. Final Result:
   Chord = 2sqrt(R^2 - r^2)

Circles Class 10 Extra Questions With Solutions

These circles class 10 extra questions cover application-based problems from CBSE-style school exams. They use only tangent, radius, and equal tangent properties.

Q1. From External Point P, Tangents PA And PB Are Drawn. If PA = 14 cm, Find PB.

PB is 14 cm. Tangents drawn from the same external point have equal lengths.

1. Given Data:
   PA = 14 cm
   PA and PB are tangents from P
2. Formula Used:
   PA = PB
3. Final Result:
   PB = 14 cm

Q2. A Circle Touches All Sides Of Quadrilateral ABCD. If AB = 8 cm, BC = 10 cm, And CD = 12 cm, Find AD.

AD is 10 cm. In a tangential quadrilateral, sums of opposite sides are equal.

1. Given Data:
   AB = 8 cm
   BC = 10 cm
   CD = 12 cm
2. Formula Used:
   AB + CD = AD + BC
3. Calculation:
   8 + 12 = AD + 10
   20 = AD + 10
   AD = 10
4. Final Result:
   AD = 10 cm

Q3. Two Tangents From P Touch A Circle At A And B. If Angle AOB = 120°, Find Angle APB.

Angle APB is 60°. The angle between tangents and angle at the centre are supplementary.

1. Given Data:
   Angle AOB = 120°
2. Formula Used:
   Angle APB + Angle AOB = 180°
3. Calculation:
   Angle APB + 120° = 180°
   Angle APB = 60°
4. Final Result:
   Angle APB = 60°

Q4. Two Tangents PA And PB Are Drawn From P. If Angle APB = 70°, Find Angle AOB.

Angle AOB is 110°. The angle between tangents and central angle are supplementary.

1. Given Data:
   Angle APB = 70°
2. Formula Used:
   Angle APB + Angle AOB = 180°
3. Calculation:
   70° + Angle AOB = 180°
   Angle AOB = 110°
4. Final Result:
   Angle AOB = 110°

Q5. A Tangent At A Meets Radius OA. What Is Angle OAP?

Angle OAP is 90°. The radius through the point of contact stands perpendicular to the tangent.

1. Given Data:
   OA is radius.
   Tangent touches the circle at A.
2. Formula Used:
   Radius ⟂ tangent
3. Final Result:
   Angle OAP = 90°

Class 10 Circles Previous Year Questions With Solutions

These class 10 circles previous year questions follow repeated board-style patterns. Each question tests one theorem through a short reasoning chain.

Q1. From P, Tangents PA And PB Are Drawn To A Circle With Centre O. If Angle PAB = 50°, Find Angle AOB.

Angle AOB is 100°. Equal tangents make triangle PAB isosceles.

1. Given Data:
   PA and PB are tangents.
   Angle PAB = 50°
2. Formula Used:
   Angle AOB + Angle APB = 180°
3. Calculation:
   PA = PB
   Angle PAB = Angle PBA = 50°
   Angle APB = 180° - 100° = 80°
   Angle AOB = 180° - 80°
4. Final Result:
   Angle AOB = 100°

Q2. O Is The Centre, AB Is A Chord, And AT Is Tangent At A. If Angle AOB = 100°, Find Angle OAT.

Angle OAT is 90°. Radius OA is perpendicular to tangent AT at A.

1. Given Data:
   O is the centre.
   AT is tangent at A.
2. Formula Used:
   Radius ⟂ tangent
3. Final Result:
   Angle OAT = 90°

Q3. AB And AC Are Tangents To A Circle With Centre O. If Angle BAC = 40°, Find Angle BOC.

Angle BOC is 140°. Radius OB and radius OC stand perpendicular to tangents AB and AC.

1. Given Data:
   Angle BAC = 40°
   AB and AC are tangents.
2. Formula Used:
   Sum of angles in quadrilateral ABOC = 360°
3. Calculation:
   Angle ABO = 90°
   Angle ACO = 90°
   Angle BOC = 360° - 90° - 90° - 40°
4. Final Result:
   Angle BOC = 140°

Q4. From External Point P, Tangents PA And PB Are Drawn. If PA = 3x + 2 And PB = 5x - 6, Find x.

The value of x is 4. Tangents drawn from the same external point are equal.

1. Given Data:
   PA = 3x + 2
   PB = 5x - 6
2. Formula Used:
   PA = PB
3. Calculation:
   3x + 2 = 5x - 6
   8 = 2x
   x = 4
4. Final Result:
   x = 4

Q5. A Point P Is 17 cm From The Centre Of A Circle. The Radius Is 8 cm. Find The Tangent Length.

The tangent length is 15 cm. The radius and tangent form a right angle.

1. Given Data:
   OP = 17 cm
   OT = 8 cm
2. Formula Used:
   PT = sqrt(OP^2 - OT^2)
3. Calculation:
   PT = sqrt(17^2 - 8^2)
   PT = sqrt(289 - 64)
   PT = sqrt(225)
4. Final Result:
   PT = 15 cm

Circles Class 10 Questions And Answers: Most Repeated Variations

These circles class 10 questions and answers cover the highest-frequency CBSE 2026 patterns. Students should practise theorem statement, proof, angle, and length variations.

Q1. What Are The Most Repeated Variations In Circles Important Questions Class 10?

The most repeated variations are radius perpendicular to tangent, equal tangents, angle between tangents, tangent length, and tangential quadrilateral.

1. Theorem 10.1:
   Use it when tangent and radius meet.
2. Theorem 10.2:
   Use it when two tangents start from one point.
3. Pythagoras Pattern:
   Use it when radius, tangent, and centre distance appear.
4. Final Result:
   These five patterns cover most Circles board questions

Q2. How Should Students Solve Class 10 Maths Chapter 10 Important Questions With Solutions?

Students should first identify the theorem, then draw the radius to the point of contact. Most errors happen when students miss the 90° angle.

1. Read The Given Data:
   Identify centre, tangent, radius, and external point.
2. Choose The Theorem:
   Use Theorem 10.1 or Theorem 10.2.
3. Apply Calculation:
   Use Pythagoras theorem for length questions.
4. Final Result:
   Correct diagram marking decides the solution path

Useful Important Questions Class 10 Maths Links