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About Circles (Tangents & Secants) — Class 10 ICSE

Study tangent properties, alternate segment theorem, and intersecting chords. This topic is part of the ICSE Class 10 mathematics syllabus (chapter: Chapter 5). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 28-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

What you'll learn in Circles (Tangents & Secants)

  • Tangents and Their Properties
  • Chord Properties and Intersecting Chords
  • Angles in Circles
  • Two Circles — Touching and Intersecting

Interactive lesson · about 25 minutes · checkpoint question after every unit

Circles (Tangents & Secants) — solved examples for Class 10 ICSE

Example 1easy

Point PP lies 2626 cm from the centre OO of a circle. The tangent PTPT drawn from PP to the circle has length 2424 cm. Find the radius.
  1. A)55 cm
  2. B)1010 cm
  3. C)1212 cm
  4. D)2525 cm

Step-by-step solution

  1. OT⊥PTOT \perp PT since the tangent meets the radius at right angles at the point of contact.
  2. Apply Pythagoras in right △OTP\triangle OTP:
    OP2=OT2+PT2OP^2 = OT^2 + PT^2
  3. Plug in the values:
    262=OT2+242  ⟹  OT2=676−576=10026^2 = OT^2 + 24^2 \implies OT^2 = 676 - 576 = 100
  4. Take the positive root:
    OT=10 cmOT = 10 \text{ cm}

Answer: 1010 cm

Example 2medium

PAPA and PBPB are tangents from external point PP to a circle with centre OO. If ∠APB=120∘\angle APB = 120^\circ and AP=5AP = 5 cm, find OPOP.
  1. A)55 cm
  2. B)525\sqrt{2} cm
  3. C)535\sqrt{3} cm
  4. D)1010 cm

Step-by-step solution

  1. OPOP is the line of symmetry, so it bisects ∠APB\angle APB:
    ∠APO=60∘\angle APO = 60^\circ
  2. OA⊥APOA \perp AP (radius ⊥\perp tangent). In right △OAP\triangle OAP:
    cos⁡60∘=APOP\cos 60^\circ = \dfrac{AP}{OP}
  3. Substitute:
    12=5OP  ⟹  OP=10 cm\dfrac{1}{2} = \dfrac{5}{OP} \implies OP = 10 \text{ cm}

Answer: 1010 cm

Example 3hard

In △ABC\triangle ABC, ∠B=90∘\angle B = 90^\circ and ABAB is a diameter of a circle that intersects ACAC at PP. The tangent at PP meets BCBC at MM. If AC=10AC = 10 cm and BC=6BC = 6 cm, find BMBM.
  1. A)22 cm
  2. B)33 cm
  3. C)44 cm
  4. D)55 cm

Step-by-step solution

  1. ABAB is a diameter, so ∠APB=90∘\angle APB = 90^\circ (angle in semicircle). Hence BP⊥ACBP \perp AC.
  2. MPMP is tangent at PP and MBMB is tangent at BB (since AB⊥BCAB \perp BC, line BCBC is tangent at BB). Tangents from external point MM are equal:
    MP=MBMP = MB
  3. ∠MPC=90∘−∠MPB=90∘−∠MBP=∠MCP\angle MPC = 90^\circ - \angle MPB = 90^\circ - \angle MBP = \angle MCP, so △MPC\triangle MPC is isosceles:
    MP=MCMP = MC
  4. Combine:
    MB=MC=BC2=62=3 cmMB = MC = \dfrac{BC}{2} = \dfrac{6}{2} = 3 \text{ cm}

Answer: 33 cm

Practice questions on Circles (Tangents & Secants)

  1. Q1.easy

    ATAT is a tangent to a circle with centre OO at the point of contact AA. If OT=4OT = 4 cm and ∠OTA=30∘\angle OTA = 30^\circ, find the length of ATAT.
    1. A)22 cm
    2. B)222\sqrt{2} cm
    3. C)232\sqrt{3} cm
    4. D)44 cm
    Show answer

    Answer: 232\sqrt{3} cm

    Hint: OA⊥ATOA \perp AT. In right △OAT\triangle OAT, use the cosine of ∠OTA\angle OTA.

  2. Q2.easy

    OO is the centre of a circle. PQPQ is a chord and PRPR is the tangent at PP. If ∠QPR=50∘\angle QPR = 50^\circ, find ∠POQ\angle POQ.
    1. A)50∘50^\circ
    2. B)80∘80^\circ
    3. C)100∘100^\circ
    4. D)130∘130^\circ
    Show answer

    Answer: 100∘100^\circ

    Hint: Tangent ⊥\perp radius gives ∠OPQ\angle OPQ. Then △OPQ\triangle OPQ is isosceles with OP=OQOP = OQ.

  3. Q3.easy

    OO is the centre of a circle, ABAB is a chord, and ATAT is the tangent at AA. If ∠AOB=100∘\angle AOB = 100^\circ, find ∠BAT\angle BAT.
    1. A)40∘40^\circ
    2. B)50∘50^\circ
    3. C)60∘60^\circ
    4. D)80∘80^\circ
    Show answer

    Answer: 50∘50^\circ

    Hint: △OAB\triangle OAB is isosceles. Then ∠BAT=90∘−∠OAB\angle BAT = 90^\circ - \angle OAB.

  4. Q4.medium

    APAP and BPBP are tangents from PP to a circle with centre OO. If AP=5AP = 5 cm and ∠APB=60∘\angle APB = 60^\circ, find the length of chord ABAB.
    1. A)2.52.5 cm
    2. B)55 cm
    3. C)525\sqrt{2} cm
    4. D)535\sqrt{3} cm
    Show answer

    Answer: 55 cm

    Hint: AP=BPAP = BP makes △PAB\triangle PAB isosceles. With ∠APB=60∘\angle APB = 60^\circ, the triangle is equilateral.

  5. Q5.medium

    PTPT and PSPS are tangents from external point PP to a circle with centre OO. If OP=2rOP = 2r where rr is the radius, find ∠TPS\angle TPS.
    1. A)30∘30^\circ
    2. B)45∘45^\circ
    3. C)60∘60^\circ
    4. D)90∘90^\circ
    Show answer

    Answer: 60∘60^\circ

    Hint: In right △OTP\triangle OTP, OT=rOT = r and OP=2rOP = 2r, so sin⁡(∠OPT)=1/2\sin(\angle OPT) = 1/2.

  6. Q6.medium

    Two equal circles with centres OO and O′O' touch each other externally at XX. ACAC is a tangent from external point AA on circle O′O' to the other circle at CC. If O′D⊥ACO'D \perp AC at DD, find the ratio O′DOC\dfrac{O'D}{OC}.
    1. A)1:31:3
    2. B)1:21:2
    3. C)2:32:3
    4. D)1:11:1
    Show answer

    Answer: 1:31:3

    Hint: O′D∥OCO'D \parallel OC (both ⊥AC\perp AC). Triangles AO′DAO'D and AOCAOC are similar — set up the ratio of corresponding sides using AO′:AOAO' : AO.

  7. Q7.hard

    △ABC\triangle ABC is right-angled at BB with AB=12AB = 12 cm and BC=5BC = 5 cm. Find the radius of the circle inscribed in the triangle.
    1. A)11 cm
    2. B)22 cm
    3. C)2.52.5 cm
    4. D)33 cm
    Show answer

    Answer: 22 cm

    Hint: For a right triangle with legs aa, bb and hypotenuse cc, the inradius is r=(a+b−c)/2r = (a + b - c)/2.

  8. Q8.hard

    Two circles with centres OO and O′O' of radii 66 cm and 33 cm touch each other internally at AA. A chord ABAB of the larger circle meets the smaller circle at CC. If AB=10AB = 10 cm, find ACAC.
    1. A)2.52.5 cm
    2. B)3.333.33 cm
    3. C)55 cm
    4. D)7.57.5 cm
    Show answer

    Answer: 55 cm

    Hint: Use that AA′AA' (extended diameters) are corresponding parts of similar triangles formed by the chord.

  9. Q9.hard

    OO is the centre of a circle with radius 55 cm. From an external point PP where OP=13OP = 13 cm, two tangents PAPA and PBPB are drawn. Find the length of PAPA.
    1. A)88 cm
    2. B)1010 cm
    3. C)1212 cm
    4. D)194\sqrt{194} cm
    Show answer

    Answer: 1212 cm

    Hint: OA⊥PAOA \perp PA, so apply Pythagoras in right △OAP\triangle OAP.

These are 9 of the 60 questions available for Circles (Tangents & Secants). Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.