Circle theorems of class 9 and class 10

 Circle theorems of class 9 and class 10  
 
 

Today in this article  I will discuss on circle theorems which are for school going students of class 9 and class 10

• Prove that perpendicular drawn from the center of a circle intersect a chord bisects the chord.

         

  Answer:-
  In the picture, a circle is drawn with center 'O'   AB is a chord of the circle. Perpendicular drawn from the center 'O' on the chord AB. i.e; OC⊥ AB 

To prove that:- OC bisects AB,i.e; AC = BC
Construction:- OA and OB are drawn.
Proof:- In between Δs AOC and BOC we get,
  ∵ OA = OB [ ∵ radii of the same circle]
  ∵ OC = OC [ ∵ Common side of the Δs]
  ∵ ∠OCA = ∠OCB [ ∵ both are 90°]
  ∴ Δ AOC ≌ Δ BOC [ By SAS congruence  rule of the Δs]
  ∴ AC = BC [∵ By CPCT]   Proved

• Prove that line joining the midpoint of a chord of a circle bisects the chord is perpendicular on the chord.

                   

Answer:-  

In the picture, AB is a chord of a circle with center 'O'  C is the midpoint of AB , OC is joined which bisects AB at point 'C'  i.e; AC = BC …..(1)
To prove:- OC ⊥ AB
Construction:- OA and OB are joined.

Proof:- In between Δs AOC and BOC we get,

  ∵ OC = OC [∵ Common side of the Δs]
   ∵ AC = BC [∵ from (1)]
and OA = OB  [∵ radii of the same circle]
∴ Δ AOC ≌ Δ BOC [∵ By SSS congruence rule of the Δs]
  ∴ ∠ OCA = <OCB [ by CPCT] but they are adjacent angles.
     ∴ ∠OCA = <OCB = 180°/2 = 90°
∴ OC⊥ AB Proved.

Prove that, angle at the center of a circle is double the angle of the circumference standing on the same arc of the circle.

          

    

Answer:- 
 In picture, <AOB and <ACB are the angles at the center and at the circumference respectively,standing on the same arc ADC of a circle with center 'O'
To prove that:- <AOB = 2<ACB
Construction:- OC is joined and extended upto D.
Proof:-  
∵ In Δ AOC, we get
               ∵  OA = OC [ ∵radii of the same 
                                   circle]
              ∴ ∠OCA = <OAC…..(1) [∵angles opposite to the equal sides are also equal]
Again,
       
          In Δ AOC
  ∵ <AOD = <OCA + <OAC [ ∵ sum of two remote interrior angles of a Δ is equal to exterior angle]
=> <AOD = <OCA+<OCA [using (1) ]
=> <AOD = 2<OCA …..(2)
Similarly we can prove,
<BOD = 2<OCB ……..(3)
Now,
     (2)+(3) we get,
<AOD+<BOD = 2(<OCA + <OCB)
=> <AOB = 2<ACB Proved.

Prove that equal chords of a circle are equidistant from the center.

              

Answer:-

  In the picture , AB and CD are two chords of a circle with center 'O' also AB=CD,and
OE⊥AB, OF⊥CD ,To prove:- OE=OF
Construction;- OA and OC are joined.
Proof:- 
∵ OE⊥ AB
 ∴ AE = ½ AB [∵ ⊥ drawn from the center on the chord bisects the chord]

Similarly,
CF= ½ CD
∵ AB = CD [∵Given]  
=>½ AB = ½ CD [∵dividing both sides by2
=>AE = CF
Now in between Δs OAE and OCF we get,
∵ AE =CF [∵ Proved]
∵ OA = OC [∵ radii of the same circle]
and, <OEA = <OFC [∵both are 90°]
  ∴ ΔOAE ≌ ΔOCF[∵by SAS rule of congruence of Δs]
∴ OE= OF [by CPCT]
      
                Proved.

    

Prove that equidistant chords of a circle are equal.

            

Answer:-

In the picture , AB and CD are two chords of a circle with center 'O'  OE⊥AB, OF⊥CD, also OE = OF, To prove:- AB=CD

 Construction:- OA and OC are joined.
Proof:- ∵ OE⊥ AB 
          ∴ AE = ½ AB [∵⊥ drawn from the center of a circle on a chord bisects the chord]
Similarly,
   
        CF = ½ CD
Now in between Δs OEA and OFC we get,

                                

   

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FREQUENTLY ASKED QUESTIONS:-

what is the measure of semi circle angle?

Answer:- Measure of semi circle angle is 90°

What is the relation between circumference angle and angle at the centre standing on the same arc of a circle?

Answer:- The relation is, Angle at the center = 2× angle of the circumference

What is the relation between all angles standing on the same arc of a circle?

Answer:- Relation is all angles standing on the same arc of a circle are equal.

What is the sum of opposite angles of a cyclic quadrilateral

Answer:- 180°

Is there any common between all the radii drawn in a circle?

Answer:- Yes, all radii drawn in a circle are equal.

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