Class 10 Maths · Chapter 10
Short answer:
Is chapter me sirf 2 exercises hain aur dono chhote hain — total karib 10-12 questions, jyada tar tangent aur radius wale 90° angle property aur Pythagoras theorem ke combination pe based. Ek theorem prove karna padta hai (tangent perpendicular to radius), baaki numericals usi property aur "do tangents from an external point equal hote hain" wale fact ko apply karke solve hote hain. Diagram-heavy chapter hai lekin concept sirf do hi hain — isliye jaldi cover ho jata hai agar dono theorems clear hon.
Circle ka tangent us point par touch karta hai jahan radius se hamesha exactly 90° banata hai — bas yahi ek fact is poore chapter ka core hai. Purane syllabus me is chapter me chords, arcs, cyclic quadrilaterals jaisa bahut kuch tha, lekin current NCERT me ye sab hata diya gaya hai. Ab sirf do cheezein bachi hain: (1) tangent radius ke perpendicular hota hai point of contact par, aur (2) kisi bhi external point se circle par exactly do tangents khinch sakte ho, aur unki length hamesha barabar hoti hai. In dono facts ko Pythagoras theorem ke saath combine karke almost saare numericals solve ho jaate hain.
Chapter 10 Summary — 5 Minute Revision
1. Tangent kya hota hai
Circle ki ek line jo circle ko sirf ek hi point par touch karti hai use tangent kehte hain, aur wo point point of contact kehlata hai. Isse compare karo secant se — secant circle ko do points par cut karta hai. Agar tumhe circle ke bahar se ek line thodi si aur andar khisaka do, wo secant ban jaati hai; jab wo line bilkul touch karke reh jaati hai to wahi tangent hai.
2. Theorem 1 — Tangent perpendicular to radius
Statement: Circle ke kisi bhi point par khincha gaya tangent, us point tak jaane wale radius ke perpendicular (90°) hota hai.
Matlab: agar circle ka centre O hai aur tangent point T par touch kar raha hai, to OT ⊥ tangent line at T — hamesha, bina exception ke. Ye fact itna reliable hai ki jab bhi kisi diagram me tangent aur radius dikh rahe hon, turant wahan 90° ka angle maan lo — chahe diagram me angle mark na ho.
Isi property ka ek seedha use: agar tangent aur ek external point diya ho, to tangent, radius, aur external-point-se-centre-ki-line milkar ek right triangle bana dete hain — aur wahin se Pythagoras aata hai.
3. Theorem 2 — External point se do tangents
Statement: Circle ke bahar kisi point P se circle par exactly do tangents khinchi ja sakti hain, aur dono tangents ki length barabar hoti hai.
Configuration yaad rakho: circle with centre O, ek external point P, dono tangents P se circle ko touch karte hain points A aur B par. To PA = PB hamesha. Isse ek symmetry bhi milti hai — OP dono tangent-lengths ka perpendicular bisector hota hai, aur △OAP ≅ △OBP (RHS congruence: OA = OB = radius, OP common, ∠OAP = ∠OBP = 90°).
4. In dono theorems ka combined use — right triangle set-up
Zyada tar numericals ka set-up ek hi tarah ka hota hai:
- Circle with centre O, radius r
- External point P, jiski distance centre O se OP di hoti hai (ya nikalni hoti hai)
- Tangent PT touching circle at point T
- Theorem 1 se: ∠OTP = 90°, isliye △OTP right-angled hai T par
- Pythagoras: OP² = OT² + PT², yaani PT² = OP² − OT² (OT = radius r)
5. Applications jo exercise me repeat hote hain
| Configuration | Kya use hota hai |
|---|---|
| Do parallel tangents ek circle ke | Unke beech ki distance = diameter (kyunki dono radius ke perpendicular hain, aur radius unke beech collinear hota hai) |
| Do tangents ek dusre ke perpendicular ho external point se | OAPB ek square ban jaata hai (jab tangent-length = radius), warna quadrilateral OAPB me angles ka sum use hota hai |
| Triangle jiske sides circle ko touch karte hain (circle inscribed) | Har vertex se nikalne wale do tangent-segments barabar hote hain — isse sides ke parts equate karke unknowns nikalte hain |
| Quadrilateral jiske sides circle ko touch karte hain | Opposite sides ke sums barabar hote hain (four tangent-length pairs ko jodne se seedha nikalta hai) |
↔ Table ko side me swipe karein
6. Common diagram vocabulary (kyunki diagrams nahi honge)
- "Circle with centre O, radius r" — basic circle
- "External point P outside the circle, OP = d" — jahan se tangent khinchni hai
- "Tangent PT touches the circle at point T" — T point of contact hai, aur OT ⊥ PT hamesha
- "PA and PB are two tangents from P touching the circle at A and B" — PA = PB, aur OP, ∠APB ko bisect karta hai
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q10)
Q1. Circle ka radius 5 cm hai aur ek point P is circle ke centre O se 13 cm door hai. P se circle par khinchi gayi tangent ki length nikalo.
Configuration: centre O, radius OT = 5 cm, external point P with OP = 13 cm, tangent PT touching circle at T.
Theorem 1 se OT ⊥ PT, isliye △OTP right-angled hai T par.
Pythagoras: OP² = OT² + PT²
13² = 5² + PT² ⇒ 169 = 25 + PT² ⇒ PT² = 144
PT = 12 cm
Q2. Ek circle ke bahar point P se do tangents PA aur PB khinchi gayi hain, jo circle ko A aur B par touch karti hain. Agar ∠APB = 80° ho, to ∠AOB kitna hoga (O centre hai)?
Configuration: OAPB quadrilateral, jahan ∠OAP = ∠OBP = 90° (Theorem 1, tangent perpendicular to radius).
Quadrilateral OAPB ke angles ka sum 360° hota hai:
∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
90° + 80° + 90° + ∠AOB = 360°
∠AOB = 360° − 260° = 100°
Q3. Circle ke centre O se, ek external point P ki tangent length PT = 24 cm hai aur OP = 25 cm hai. Circle ka radius nikalo.
Configuration: OT ⊥ PT (Theorem 1), so △OTP right-angled at T.
OP² = OT² + PT²
25² = OT² + 24² ⇒ 625 = OT² + 576
OT² = 49 ⇒ OT = 7 cm
Radius = 7 cm.
Q4. Prove karo: circle ke kisi bhi point par khincha gaya tangent, us point tak radius ke perpendicular hota hai.
Ye khud Theorem 1 ka proof hai (contradiction method se):
Maan lo circle with centre O, aur ek tangent line l point T par touch karti hai (T point of contact hai).
Assume ki OT, l ke perpendicular NAHI hai. To O se l par ek dusra point Q lo jahan OQ ⊥ l ho (perpendicular distance sabse chhota hota hai).
Right triangle OQT me, hypotenuse OT hoga (kyunki ∠OQT = 90°), isliye OT > OQ.
Lekin l tangent hai, matlab l circle ko sirf T par touch karti hai — koi aur point l par circle ke andar/circle par nahi ho sakta.
Agar OQ < OT = radius, to Q circle ke andar aa jaata (kyunki OQ < radius), jo l ke tangent hone ke against hai — l to sirf T par touch karti hai, kaheen aur circle ko cross ya touch nahi karti.
Ye contradiction dikhata hai ki assumption galat thi.
Isliye OT hi l par sabse chhota perpendicular distance hai, matlab OT ⊥ l. Hence proved — tangent radius ke perpendicular hota hai point of contact par.
Q5. Ek external point P se circle par do tangents PA aur PB khinchi gayi hain jo A aur B par touch karti hain. Agar PA = 10 cm ho, to PB kitna hoga? Reasoning bhi do.
Theorem 2 (external point se do tangents) seedha yahan apply hota hai:
Ek hi external point se circle par khinche gaye do tangents ki length hamesha barabar hoti hai.
Isliye PB = PA = 10 cm.
Reason: △OAP aur △OBP RHS se congruent hain — OA = OB (radius), OP common, ∠OAP = ∠OBP = 90° (Theorem 1). Isse PA = PB milta hai.
Q6. Ek circle ke centre O se do parallel tangents khinchi gayi hain, jo circle ko A aur B par touch karti hain. Prove karo ki AB circle ka diameter hai.
Configuration: circle with centre O, tangent l₁ touching at A, tangent l₂ (parallel to l₁) touching at B.
Theorem 1 se OA ⊥ l₁ aur OB ⊥ l₂.
Kyunki l₁ ∥ l₂, aur OA, OB dono unke perpendicular hain, isliye OA aur OB dono ek hi line par honge (ek common perpendicular direction).
Matlab O, A aur B teenon collinear hain, aur A, B circle ke opposite sides par hain (kyunki l₁, l₂ parallel tangents hain, circle ke do alag taraf).
Isliye AB, O se guzarta hua ek chord hai jiski dono ends circle par hain aur O uske beech me hai — yahi diameter ki definition hai. Hence AB diameter hai.
Q7. Ek right triangle jiska right angle B par hai, uske do sides AB = 6 cm aur BC = 8 cm hain. Is triangle ke andar ek circle (incircle) hai jiska centre O aur radius r hai, jo teeno sides ko touch karta hai. r nikalo. (Hint: tangent lengths use karo)
Configuration: incircle touches AB at P, BC at Q, CA at R. Kyunki B par right angle hai aur BPOQ ke sides OP ⊥ AB, OQ ⊥ BC (Theorem 1), aur OP = OQ = r, isliye BPOQ ek square hai side r ka.
Isliye BP = BQ = r.
Tangent lengths equal (Theorem 2): AP = AR, CQ = CR.
AB = AP + PB = AP + r ⇒ AP = 6 − r
BC = BQ + QC = r + QC ⇒ QC = 8 − r
AC = √(AB² + BC²) = √(36 + 64) = √100 = 10 cm
AC = AR + RC = AP + QC (kyunki AR = AP, RC = QC)
10 = (6 − r) + (8 − r) = 14 − 2r
2r = 4 ⇒ r = 2 cm
Q8. Ek quadrilateral ABCD circle ko circumscribe karta hai (matlab circle uske chaaron sides ko touch karta hai). Prove karo: AB + CD = BC + AD.
Configuration: incircle touches AB at P, BC at Q, CD at R, DA at S.
Tangent lengths from each vertex equal hote hain (Theorem 2):
AP = AS, BP = BQ, CQ = CR, DR = DS
AB + CD = (AP + PB) + (CR + RD)
= (AS + BQ) + (CQ + DS) [substituting equal tangent pairs]
= (AS + DS) + (BQ + CQ)
= AD + BC
Isliye AB + CD = BC + AD. Hence proved.
Q9. Ek external point P se circle (centre O, radius 5 cm) par do tangents PA aur PB is tarah khinchi gayi hain ki ∠APB = 60° hai. PA ki length nikalo.
Configuration: OAPB quadrilateral, OP diagonal ∠APB ko bisect karta hai (symmetry se, kyunki △OAP ≅ △OBP).
∠APO = ∠APB / 2 = 30°
△OAP right-angled hai A par (Theorem 1: OA ⊥ PA)
tan(∠APO) = OA / PA
tan 30° = 5 / PA ⇒ PA = 5 / tan 30° = 5 / (1/√3) = 5√3 cm
Q10. Do circles ek dusre ko bahar se touch karte hain (externally touching), radii 4 cm aur 6 cm. In dono circles ki ek common tangent PQ hai jahan P pehle circle par aur Q dusre circle par hai, aur ye tangent unke centres ko milane wali line se X par milti hai jaha X touch point hai. Agar PQ 4 cm aur 6 cm radius wale circles ke beech ka ek direct common tangent hai (centres O₁O₂ ki distance 10 cm), to PQ ki length nikalo.
Configuration: centre O₁ (radius 4), centre O₂ (radius 6), O₁O₂ = 10 cm (kyunki circles bahar se touch karte hain: 4 + 6 = 10). PQ direct common tangent hai, touching first circle at P and second at Q.
Direct common tangent ki length ka formula: PQ = √(O₁O₂² − (r₁ − r₂)²)
Ye formula bhi Theorem 1 se hi nikalta hai — O₁ se O₂Q ke parallel ek line kheenchne par ek right triangle banta hai jiska ek side (r₂ − r₁) aur hypotenuse O₁O₂ hota hai.
PQ = √(10² − (6 − 4)²) = √(100 − 4) = √96 = 4√6 cm
Important Equations — Ek Nazar Me
| Theorem / Formula | Statement |
|---|---|
| Theorem 1 — Tangent ⊥ Radius | Circle ke kisi point T par tangent, us point tak radius OT ke perpendicular hota hai: ∠OTP = 90° (P kisi bhi point ho tangent line par) |
| Theorem 2 — Equal tangents from external point | External point P se circle par exactly 2 tangents banti hain, PA aur PB, aur PA = PB |
| Tangent-length formula (Pythagoras se) | PT² = OP² − OT², jahan O centre, P external point, T point of contact, OT = radius |
| Parallel tangents ka distance | Do parallel tangents ke beech ki distance = diameter = 2r |
| Circle inscribed in quadrilateral | Opposite sides ke sums barabar: AB + CD = BC + AD |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Angle ko 90° maan kar bhi likhna bhool jaana. Diagram me agar tangent aur radius dono diye hain, turant 90° maan lo aur solution me explicitly likho "∠OTP = 90° (tangent ⊥ radius)" — bina justification ke marks kat sakte hain.
- Pythagoras me galat sides lagana. Hamesha yaad rakho hypotenuse OP hai (centre se external point), aur dusre do sides radius (OT) aur tangent-length (PT) hain. Kabhi bhi tangent ko hypotenuse maan kar formula ulta mat likho.
- Tangent-length ko bina radius-perpendicular property justify kiye seedha formula laga dena. PT² = OP² − OT² sirf isliye valid hai kyunki OT ⊥ PT hai — ye step likhna zaroori hai, warna proof incomplete maana jaata hai.
- Chord aur tangent ko confuse karna. Chord circle ke do points ko andar se jodta hai, tangent sirf ek point ko touch karta hai bahar se. Dono ke properties bilkul alag hain — chord ke perpendicular bisector wali property yahan is chapter me use nahi hoti.
- External point se do tangents ki length automatically barabar samajh kar bina reason diye likhna. Jab tak proof nahi maanga jaye tab tak seedha use karna theek hai, lekin agar "prove karo" bola ho to RHS congruence (△OAP ≅ △OBP) dikhana zaroori hai.
- Quadrilateral OAPB ka angle-sum bhool jaana. Jab O, A, P, B se banta quadrilateral ho (do radius aur do tangents), uske angles ka sum 360° hota hai — ye formula bhool jaane se ∠AOB aur ∠APB wale questions atak jaate hain.
Board-Style Important Questions
- 1 mark: Ek circle ke centre O se ek external point P ki doori 10 cm hai aur circle ka radius 6 cm hai. P se circle par khinchi gayi tangent ki length kya hogi?
- 1 mark: Tangent ka circle ke radius ke saath kya angle banta hai point of contact par?
- 2 marks: Do tangents PA aur PB ek external point P se circle par khinchi gayi hain, A aur B touch points hain. Agar ∠AOB = 130° ho (O centre), to ∠APB nikalo.
- 3 marks: Prove karo ki circle ke kisi point par khincha gaya tangent, us point tak radius ke perpendicular hota hai.
- 5 marks: Ek quadrilateral ABCD ek circle ko circumscribe karta hai. Prove karo ki AB + CD = BC + AD, aur clearly batao tangent-length property kaise use hui.
Quick Quiz — Score Check Karein
Q1. Class 10 Maths Chapter 10 'Circles' me is chapter ka core fact kya hai?
Q2. Tangent aur secant me kya fark hai?
Q3. Circle ka radius 5 cm hai aur external point P, centre O se 13 cm door hai. P se tangent ki length kitni hogi?
Q4. Ek circle par external point P se tangent PT ki length 24 cm hai aur OP = 25 cm hai. Circle ka radius kya hoga?
Q5. Theorem 2 ke according, ek external point P se circle par kitni tangents khinchi ja sakti hain, aur unki length ka kya relation hota hai?
Q6. External point P se do tangents PA aur PB circle ko A aur B par touch karti hain. Agar ∠APB = 80° ho, to ∠AOB (O centre) kitna hoga?
Q7. Quadrilateral OAPB me (O centre, P external point, A aur B tangent points) ∠OAP aur ∠OBP kitne hote hain?
Q8. Do parallel tangents ek circle ke beech ki distance kya hoti hai?
Q9. Agar ek quadrilateral ke saare sides circle ko touch karte hain (circumscribed quadrilateral), to kya property hoti hai?
Q10. Theorem 1 (tangent perpendicular to radius) ka proof kis method se kiya jaata hai jaisa chapter me diya gaya hai?
Aksar Poochhe Jaane Wale Sawaal
Circle aur tangent chapter me chords, cyclic quadrilaterals wale theorems kyun nahi hain?
Current rationalised NCERT syllabus me ye chapter sirf tangent properties tak trim kar diya gaya hai — chords/arcs/cyclic quadrilateral wale topics is chapter se hata diye gaye hain, isliye humne bhi wahi scope follow kiya hai.
Tangent aur radius ka angle hamesha exactly 90° hota hai ya approximately?
Exactly 90° — ye ek proven theorem hai, koi approximation nahi. Har diagram me chahe angle mark ho ya na ho, ye fact hamesha true hota hai.
External point se 3 ya usse zyada tangents kyun nahi ban sakte?
Geometrically circle ke bahar kisi bhi point se sirf 2 hi lines aisi ho sakti hain jo circle ko exactly ek point par touch karein — teesri koi bhi line ya to circle ko cross karegi (secant) ya bilkul miss karegi.
Agar point circle ke andar ho to tangent kaise banega?
Circle ke andar wale point se koi tangent nahi banta — tangent sirf circle ke bahar ke point se ya khud circle ke point se banta hai. Andar wale point se har line circle ko do jagah cross karegi (secant).
PT² = OP² − OT² formula yaad rakhna chahiye ya derive karna chahiye har baar?
Derive karna better hai — pehle likho "OT ⊥ PT (tangent ⊥ radius)", phir Pythagoras apply karo. Bina derivation ke seedha formula lagana kabhi kabhi step marks kata deta hai.
Do circles ke common tangent wale questions is chapter me aate hain kya?
Basic direct/transverse common tangent length nikalna is chapter ke core theorems (tangent ⊥ radius) se hi derive hota hai, isliye occasionally aise applied questions aa sakte hain — lekin inka base wahi do theorems hain jo humne cover kiye.
Class 10 Maths — Saare Chapters
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