CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 9 Circles
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Boards begin February 2027
Pre-boards are in December
Circles is in Geometry, which carries 15 of the 80 marks in the 2026-27 CBSE Maths paper
14 NCERT Exemplar questions from Chapter 9
Exercise 9.1
Exemplar Ex 9.1 Q11 mark
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is
Step 1Let AB be a chord of the larger circle tangent to the smaller circle at P. OP ⊥ AB, OP = 4.
Step-by-step solution
Let AB be a chord of the larger circle tangent to the smaller circle at P. OP ⊥ AB, OP = 4.
OA = 5. In right ∆OPA: AP² = 25 − 16 = 9 ⇒ AP = 3.
From a point P at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR are drawn. The area of the quadrilateral PQOR is
Step 1PQ = √(13² − 5²) = √144 = 12 cm. Similarly PR = 12 cm.
Step-by-step solution
PQ = √(13² − 5²) = √144 = 12 cm. Similarly PR = 12 cm.
Each of ∆PQO and ∆PRO is right-angled with legs 5 and 12.
Area of one such triangle = (1/2) × 5 × 12 = 30 cm².
Is it true that the length of the tangent from an external point on a circle is always greater than the radius of the circle? Justify with a counter-example.
Step 1Take a circle of radius 5 and an external point P with OP = 6. Tangent length = √(36 − 25) = √11 ≈ 3.32, which is LESS than radius 5.
Step-by-step solution
Take a circle of radius 5 and an external point P with OP = 6. Tangent length = √(36 − 25) = √11 ≈ 3.32, which is LESS than radius 5.
Hence the statement is FALSE.
Answer:False; counter-example r = 5, OP = 6 gives tangent ≈ 3.32 < 5.
Out of two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
Step 1Let P be the point of contact on the inner circle. OP ⊥ AC and bisects AC at P, so AP = 4 cm.
Step-by-step solution
Let P be the point of contact on the inner circle. OP ⊥ AC and bisects AC at P, so AP = 4 cm.
From an external point P, two tangents PA and PB are drawn to a circle with centre O. At a point E on the circle, the tangent intersects PA at C and PB at D. If PA = 10 cm, find the perimeter of triangle PCD.
Step 1[1 mark] Tangents drawn from an external point are equal, so PA = PB = 10 cm.
Step-by-step solution
[1 mark] Tangents drawn from an external point are equal, so PA = PB = 10 cm.
[1 mark] From C the two tangents give CA = CE; from D they give DB = DE (equal tangents from an external point).
[1 mark] Perimeter(△PCD) = PC + CD + DP = PC + (CE + ED) + DP.
[1 mark] Substitute CE = CA and ED = DB: = (PC + CA) + (DP + DB) = PA + PB.
[1 mark] Perimeter = 10 + 10 = 20 cm. Hence the perimeter of △PCD is 20 cm.
AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD.
Step 1∠ACB = 90° (angle in a semicircle). In ∆ABC: ∠ABC = 180° − 90° − 30° = 60°.
Step-by-step solution
∠ACB = 90° (angle in a semicircle). In ∆ABC: ∠ABC = 180° − 90° − 30° = 60°.
OC ⊥ CD (tangent ⊥ radius), so ∠OCD = 90°.
∠OCB = ∠OBC (isosceles, OC = OB = radius), and ∠OBC = 60°. So ∠OCB = 60°.
∠BCD = ∠OCD − ∠OCB = 90° − 60° = 30°.
Exterior angle at B in ∆BCD: ∠OBC = 60° = ∠BDC + ∠BCD ⇒ ∠BDC = 60° − 30° = 30°.
Thus ∠BCD = ∠BDC = 30° ⇒ ∆BCD is isosceles ⇒ BC = BD.
If d₁ and d₂ (with d₂ > d₁) are the diameters of two concentric circles and c is the length of a chord of the larger circle that is tangent to the smaller circle, prove that d₂² = c² + d₁².
Step 1Let the chord AB of the larger circle touch the smaller circle at C. Then OC ⊥ AB and OC = d₁/2.
Step-by-step solution
Let the chord AB of the larger circle touch the smaller circle at C. Then OC ⊥ AB and OC = d₁/2.
OA = d₂/2 (radius of larger). C is the midpoint of AB, so CB = c/2.
In right ∆OCB: OC² + CB² = OB² ⇒ (d₁/2)² + (c/2)² = (d₂/2)².
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Assertion-Reason1 mark
Assertion (A): If the angle between two tangents drawn from an external point P to a circle of radius a is 90°, then OP = a√2.
Reason (R): In the right triangle formed by O, the point of contact and P, the angle at P is half the angle between the tangents.
Step 1By R, the angle at P in the half-triangle is 45°. So sin 45° = a/OP ⇒ OP = a/sin 45° = a√2 — matches A.
Step-by-step solution
By R, the angle at P in the half-triangle is 45°. So sin 45° = a/OP ⇒ OP = a/sin 45° = a√2 — matches A.
R is the standard 'OP bisects angle between tangents' result — true.
A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are tangents to the circle at P and Q. A tangent BC drawn at a point R on the minor arc PQ intersects AP at B and AQ at C.
(i) Find AP.
(ii) Are BP and BR equal? Why?
(iii) Are CQ and CR equal? Why?
(iv) Find the perimeter of ∆ABC.
Step 1(i) In right ∆OPA: AP² = OA² − OP² = 169 − 25 = 144 ⇒ AP = 12 cm.
Step-by-step solution
(i) In right ∆OPA: AP² = OA² − OP² = 169 − 25 = 144 ⇒ AP = 12 cm.
(ii) BP and BR are tangents from the same external point B ⇒ BP = BR.
(iii) Similarly CQ = CR.
(iv) Perimeter(ABC) = AB + BC + CA = AB + (BR + RC) + CA = AB + BP + CQ + CA = (AB + BP) + (CA + CQ) = AP + AQ = 12 + 12 = 24 cm.
Answer:(i) AP = 12 cm; (ii) Yes (equal tangents from B); (iii) Yes (equal tangents from C); (iv) Perimeter = 24 cm.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.