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Subject Mathematics
Class IX
•TOPIC
CIRCLES
09/11/1805:07AMHARISHKUMAR
1
Angle subtended by a chord at centre of
a circle
09/11/1805:07AMHARISHKUMAR
2
Equal chords of a circle subtends equal
angle at centre
• OA = OD (WHY ?)
• OB = OC (WHY ?)
• AB = CD (WHY ?)
• ∆AOB ≅ ∆COD (SSS)
• ∠AOB = ∠COD
09/11/1805:07AMHARISHKUMAR
3
O
BC
D
A
If angles subtended by chords of a circle
at centre are equal then the chords are
equal
• OA = OD (WHY ?)
• OB = OC (WHY ?)
• ∠AOB = ∠COD (WHY
?)
• ∆AOB≅ ∆COD (SAS)
• AB = CD
09/11/1805:07AMHARISHKUMAR
4
O
BC
D
A
The perpendicular from centre of a
circle to a chord bisects the chord
• OA=OB (WHY ?)
• ∠ OCA=∠OCB
(WHY ?)
• OC=OC (WHY ?)
• ∆ OAC ∆ ≅OBC
(WHY ?)
• AC=BC
09/11/1805:07AMHARISHKUMAR
5
O
A
B
C
Line segment joining centre to mid
point of a chord is perpendicular to the
chord
• OA=OB (WHY ?)
• AC=BC (WHY ?)
• OC=OC (WHY ?)
• ∆OAC≅ ∆OBC
(WHY ?)
• ∠ACO= ∠BCO=90
09/11/1805:07AMHARISHKUMAR
6
O
A
B
C
Circle through three points
• OA=OB
• OB=OC
• OA=OB=OC
09/11/1805:07AMHARISHKUMAR
7
A
B
C
O
SUMMARY
• Equal chords subtend equal angle at centre
• If angle subtended by chords at centre are equal
then chords are equal
• Perpendicular from centre bisects the chord
• Line segment joining mid point of a chord and
centre of circle is perpendicular to the chord
• A circle can be drawn through any three Non
collinear points
09/11/1805:07AMHARISHKUMAR
8
• If AB = CD
• Then
• ∠AOB = ∠COD
09/11/1805:07AMHARISHKUMAR
9
O
BC
D
A
• If ∠AOB = ∠COD
• Then
• AB = CD
09/11/1805:07AMHARISHKUMAR
10
O
BC
D
A
• If OC ⊥ AB
• Then
• AC=BC
09/11/1805:07AMHARISHKUMAR
11
O
A
B
C
• IF AC=BC
• THEN
• OC ⊥ AB
09/11/1805:07AMHARISHKUMAR
12
O
A
B
C
• Q. From an externl
point P, a tangent PT
and a line segment
PAB is drawn to a
circle with centre O.
ON is perpendicular
on the chord AB.
• Prove that :
• PA.PB = PT2
13
HARISHKUMAR09/11/1805:07AM
• Q. If a circle touches
the side BC of a
triangle ABC at P and
extended sides AB
and AC at Q and
R.Prove that
• AQ = ½(BC+CA+AB)
14
HARISHKUMAR09/11/1805:07AM
• Q. If a,b,c are the
sides of a right
triangle where c is
the hypotenuse
prove that the radius
r of the circle which
touches the sides of
the triangle is given
by r = (a+b-c)/2
15
HARISHKUMAR09/11/1805:07AM
• Q. Angle in alternate
segment.
16
HARISHKUMAR09/11/1805:07AM
• Q. AB is a diameter of a circle and chord CD = radius OC. If AC
and BD when produced meet at P prove that angle APB = 60
17
HARISHKUMAR09/11/1805:07AM
• Q. Prove that altitudes of a triangle are concurrent.
18
HARISHKUMAR09/11/1805:07AM

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CLASS IX MATHS

  • 2. Angle subtended by a chord at centre of a circle 09/11/1805:07AMHARISHKUMAR 2
  • 3. Equal chords of a circle subtends equal angle at centre • OA = OD (WHY ?) • OB = OC (WHY ?) • AB = CD (WHY ?) • ∆AOB ≅ ∆COD (SSS) • ∠AOB = ∠COD 09/11/1805:07AMHARISHKUMAR 3 O BC D A
  • 4. If angles subtended by chords of a circle at centre are equal then the chords are equal • OA = OD (WHY ?) • OB = OC (WHY ?) • ∠AOB = ∠COD (WHY ?) • ∆AOB≅ ∆COD (SAS) • AB = CD 09/11/1805:07AMHARISHKUMAR 4 O BC D A
  • 5. The perpendicular from centre of a circle to a chord bisects the chord • OA=OB (WHY ?) • ∠ OCA=∠OCB (WHY ?) • OC=OC (WHY ?) • ∆ OAC ∆ ≅OBC (WHY ?) • AC=BC 09/11/1805:07AMHARISHKUMAR 5 O A B C
  • 6. Line segment joining centre to mid point of a chord is perpendicular to the chord • OA=OB (WHY ?) • AC=BC (WHY ?) • OC=OC (WHY ?) • ∆OAC≅ ∆OBC (WHY ?) • ∠ACO= ∠BCO=90 09/11/1805:07AMHARISHKUMAR 6 O A B C
  • 7. Circle through three points • OA=OB • OB=OC • OA=OB=OC 09/11/1805:07AMHARISHKUMAR 7 A B C O
  • 8. SUMMARY • Equal chords subtend equal angle at centre • If angle subtended by chords at centre are equal then chords are equal • Perpendicular from centre bisects the chord • Line segment joining mid point of a chord and centre of circle is perpendicular to the chord • A circle can be drawn through any three Non collinear points 09/11/1805:07AMHARISHKUMAR 8
  • 9. • If AB = CD • Then • ∠AOB = ∠COD 09/11/1805:07AMHARISHKUMAR 9 O BC D A
  • 10. • If ∠AOB = ∠COD • Then • AB = CD 09/11/1805:07AMHARISHKUMAR 10 O BC D A
  • 11. • If OC ⊥ AB • Then • AC=BC 09/11/1805:07AMHARISHKUMAR 11 O A B C
  • 12. • IF AC=BC • THEN • OC ⊥ AB 09/11/1805:07AMHARISHKUMAR 12 O A B C
  • 13. • Q. From an externl point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. • Prove that : • PA.PB = PT2 13 HARISHKUMAR09/11/1805:07AM
  • 14. • Q. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R.Prove that • AQ = ½(BC+CA+AB) 14 HARISHKUMAR09/11/1805:07AM
  • 15. • Q. If a,b,c are the sides of a right triangle where c is the hypotenuse prove that the radius r of the circle which touches the sides of the triangle is given by r = (a+b-c)/2 15 HARISHKUMAR09/11/1805:07AM
  • 16. • Q. Angle in alternate segment. 16 HARISHKUMAR09/11/1805:07AM
  • 17. • Q. AB is a diameter of a circle and chord CD = radius OC. If AC and BD when produced meet at P prove that angle APB = 60 17 HARISHKUMAR09/11/1805:07AM
  • 18. • Q. Prove that altitudes of a triangle are concurrent. 18 HARISHKUMAR09/11/1805:07AM