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Circles

Lesson 18 of 113

1. Ye topic hai kya?

Circle = saare points jo centre O se equal doori r (radius). Class-X ka kaam: tangent — line jo circle ko ek hi point pe chhu-ti. Paper: radius tangent se 90°, bahar ke point se do tangents barabar, OP² = r² + (tangent)², angle APB se angle AOB.

Yahan lock

Touch point, ⊥ radius, 0/1/2 tangents, PA=PB, OP symmetry, OAPB cyclic (90°+90°), kite, incircle se vertex lengths.

Yahan kya nahi

Sector area, 22/7 se plot. Chord-angle Class-IX cyclic (angle in semicircle) yahan theorem nahi. Compass construction. Tower 30°. Coordinate x²+y²=r² equation XI.

2. Tangent vs secant

Secant circle ko do points pe kaat-ti. Tangent ek point — point of contact. Us point pe circle aur line same direction (locally).

Point PTangentsSocho
Andar (OP < r)0Koi line do baar kaat-ti ya andar
Circle pe (OP = r)1Wahi tangent at P
Bahar (OP > r)2Do touch points A, B
Chord ka extend secant ho sakta; chord khud tangent nahi. Diameter bhi secant family. “Circle pe point se infinite tangents” galat — exactly one.

3. Radius ⊥ tangent (touch pe)

Tangent at A, centre O: OA ⊥ tangent. Matlab triangle OAP (P tangent pe koi dusra point) right-angled at A. Length: PA = √(OP² − r²).
O A r tangent

OA radius, A touch. Square mark = 90°. Tangent A ke dono taraf jaati; ⊥ sirf A pe.

Kyun: O se tangent tak shortest distance perpendicular hai, aur shortest = radius (warna line circle ko do baar kaat-ti — contradiction).

Example

r = 7, P bahar, tangent PA = 24. OP?

Solution

  1. Right at A

    OA ⊥ PA. 7² + 24² = OP² → 49 + 576 = 625. OP = 25.

  2. Nahi

    7+24=31. Tower/sihi 13-5 yahan nahi — yahan radius aur tangent legs.

Answer25

4. Bahar se do tangents — PA = PB

P se do tangents, touch A aur B. OA ⊥ PA, OB ⊥ PB, OA = OB = r.

PA = PB. Triangles OAP aur OBP congruent (HL: hypotenuse OP common, leg r). Isliye angles bhi: OP bisect karta ∠APB aur ∠AOB. Kite OAPB: OA=OB, PA=PB.
O A B P PA PB

PA = PB. Dashed OP symmetry. 90° at A and at B.

Example

OP = 15, r = 9. Har tangent ki length?

Solution

  1. √(OP²−r²)

    √(225−81)=√144=12. Dono 12.

Answer12

5. Angles — OAPB

Quadrilateral OAPB: ∠OAP = ∠OBP = 90°. Opposite angles sum: 90°+90° + ∠AOB + ∠APB = 360° →

∠AOB + ∠APB = 180°. OP bisects dono. Triangle OAP mein: ∠OAP=90°, ∠APO = ½∠APB, ∠AOP = ½∠AOB.
Example

Do tangents, ∠APB = 80°. ∠AOB?

Solution

  1. 180

    ∠AOB = 180° − 80° = 100°. Phir OP se 50° + 50°.

Answer100°

Special: ∠APB = 90° → ∠AOB = 90°, PA = r (45°-45°-90° in OAP), OP = r√2.

OAPB cyclic isliye ke opposite 90°. Circle theorems IX (same arc, cyclic quad inscribed) alag — yahan sirf yeh tangent-quad.

6. Incircle — equal tangents se sides

Triangle ke andar circle, teen sides ko touch. Har vertex se do tangents equal. Sides a, b, c, s = (a+b+c)/2. Vertex A se tangent length = s − a.

Example

Triangle sides 13, 14, 15. Incircle. Vertex opposite 15 se tangent length?

Solution

  1. s

    s = 21. Opposite 15 wali side a=15, length s−a = 6.

  2. Check

    Teen lengths 6, 7, 8: 6+7=13, 6+8=14, 7+8=15. (Compass se draw Constructions pe.)

Answer6

7. Sawal — basic se pro

Q1 · Basic

Tangent circle ko kitne points pe?

  1. 1
  2. 2
  3. 0
  4. Infinite

Solution

  1. Touch

    1. B secant.

Answer1
Q2 · Basic

Tangent at A, OA kya?

  1. Tangent ke ⊥
  2. Tangent ke parallel
  3. Chord AB ke barabar
  4. Hamesha 45°

Solution

  1. Theorem

    ⊥ at touch.

AnswerA
Q3 · Basic

r=7, PA=24. OP?

  1. 25
  2. 31
  3. 17
  4. √24

Solution

  1. 49+576

    625, 25. B sum. C 8-15 mix.

Answer25
Q4 · Basic

P circle ke andar. Tangents?

  1. 0
  2. 1
  3. 2
  4. Infinite

Solution

  1. OP<r

    0.

Answer0
Q5 · Medium

Do tangents P se. PA = 12. PB?

  1. 12
  2. 6
  3. 24
  4. Nahi pata r ke bina

Solution

  1. Equal

    PA=PB hamesha, r ki zaroorat nahi.

Answer12
Q6 · Medium

OP=15, r=9. Tangent length?

  1. 12
  2. 6
  3. √15
  4. 24

Solution

  1. √144

    12. B 15−9.

Answer12
Q7 · Medium

∠APB = 80°. ∠AOB?

  1. 100°
  2. 80°
  3. 40°
  4. 180°

Solution

  1. 180−80

    100°. C: half OP wala, sawal AOB poochha.

Answer100°
Q8 · Medium

∠OAP (tangent at A, radius OA)?

  1. 90°
  2. 60°
  3. 45° hamesha
  4. 0°

Solution

  1. ⊥

    90°. C tab jab APB=90 special.

Answer90°
Q9 · Pro

Sides 13, 14, 15, incircle. Opposite 15 se tangent?

  1. 6
  2. 8
  3. 7
  4. 21

Solution

  1. s−a

    21−15=6. B opposite 13. C opposite 14. D s.

Answer6
Q10 · Pro

∠APB = 90°, r = 6. OP?

  1. 6√2
  2. 6
  3. 12
  4. 3√2

Solution

  1. AOB=90

    OAP: 90-45-45, PA=r=6, OP=6√2.

Answer6√2

Practice

Pehle khud, phir neeche kholo.

12345 678910

P1

Point circle pe. Tangents?

  1. 1
  2. 2
  3. 0
  4. Infinite

Solution

  1. OP=r

    Exactly one.

Answer1
P2

OP=13, r=5. Tangent?

  1. 12
  2. 8
  3. 18
  4. √13

Solution

  1. 169−25

    144, 12. (5-12-13 yahan OP hyp, r aur tangent legs — ladder 13 m alag chapter.)

Answer12
P3

OP kya bisect karta?

  1. ∠APB aur ∠AOB dono
  2. Sirf chord AB
  3. Kuch nahi
  4. Sirf tangent PA

Solution

  1. Congruent

    Dono angles. AB bhi ⊥ OP pe, lekin option B “sirf chord” adhoora.

AnswerA
P4

∠APB=60°. ∠AOB?

  1. 120°
  2. 60°
  3. 30°
  4. 90°

Solution

  1. 180−60

    120°.

Answer120°
P5

Secant kitne points?

  1. 2
  2. 1
  3. 0
  4. 3

Solution

  1. Cut

    2. Tangent 1.

Answer2
P6

OAPB mein ∠OAP + ∠OBP?

  1. 180°
  2. 90°
  3. 360°
  4. Depends on r

Solution

  1. 90+90

    180. Isliye cyclic / AOB+APB=180.

Answer180°
P7

r=9, tangent=12. OP?

  1. 15
  2. 21
  3. √12
  4. 3

Solution

  1. 81+144

    225, 15.

Answer15
P8

Incircle, sides 13,14,15. Opposite 13 se length?

  1. 8
  2. 6
  3. 7
  4. 14

Solution

  1. 21−13

    8.

Answer8
P9

PA tangent, P circle ke bahar, OA radius. Triangle OAP right kahan?

  1. A
  2. O
  3. P
  4. Kabhi nahi

Solution

  1. Touch

    Right at A, nahi O (O pe two radii angle), nahi P.

AnswerA
P10

∠APB=90°, r=6. Tangent PA?

  1. 6
  2. 6√2
  3. 3
  4. 12

Solution

  1. Isosceles right

    PA=OA=6. B OP hai.

Answer6

← Index · Applications of trigonometry

Agle topic: Constructions.