Class IX circle formula

                                                                          CIRCLES

                                                                 CLASS-IX 

Circle

A circle is a set of those points in a plane which are at given

distance from a given fixed point on the plane. The fixed point is                     📀

called centre and fixed distance is called radius of the circle.

Arc of a Circle

A continuous piece of the circle is called arc of a circle. There are

two types of arcs : minor arc and major arc. Arc AB is minor arc,                              ⌒

since its length is less than half of the perimeter. The length of this

arc is written as m( AB)

Chord and Segment of a Circle

A line joining any two points on the circle is called a chord of the                             ⌽

circle.

Minor Segment: The smaller part of the circle divided by a chord

AB is called minor segment of the circle.

Major Segment: The bigger part of the circle is called major segment of the circle.

Cyclic Quadrilateral

A quadrilateral is called cyclic, if all its vertices lie on a circle. The sum of opposite angles of a cyclic

quadrilateral is always 180°.

                                                            THEOREMS

Theorem 1 : The perpendicular from the centre of a circle to a chord bisects the chord.

Given: A chord PQ of a circle C(O,r) and perpendicular OL to the chord PQ.

To Prove LP= LQ

Construction   Join OP and OQ.

Proof  In triangle PLO and QLO, we have

  OP= OQ= r   [Radii of the same circle]

  OL= OL        [Common]

& ∠OLP=∠ OLQ   [Each equal to 90°]                                                 ⍥

So, by RHS - criterion of congruence, we have

⊿PLO =⊿ QLO

PL=LQ

Theorem 2 : (Converse of above theorem) The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.

Given   A chord PQ of a circle C(O, r) with mid-point M.

To Prove   OM pependiculur to  PQ.

Construction   Join OP and OQ

Proof      In triangle OPM and OQM, we have

               OP= OQ                       [Radii of the same circle]

              PM = MQ                     [M is mid-point of PQ]

             OM=OM                       [Common]

           So, by SSS-criterion of congruence, we have

          ΔΟΡΜ = ΔΟΜ

         ∠OMP= ∠OMQ

         But,∠ OMP+ ∠OMQ=180°      [Linear pair]

        ∠OMP+ ∠OMP=180°               [: 2OMP = 20MQ]

2∠OMP=180°

∠OMP=90°

O∠MP= ∠OMQ=90°   [:∠ OMP= ∠OMQ]

Hence, OM perpendicular to  PQ.

Theorem 3 : Equal chords of a circle are equidistant from the centre.

Given      Two chords AB and CD of circle C(O, r) such that AB = CD and OL perpendicular to  AB and OM perpendicular to  CD.

To Prove    Chords AB and CD are equidistant from the centre O i.e., OL = OM.

Construction  Join OA and OC.

Proof     Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,

            OL perpendicular to AB  = AL=1/2 AB   .... (i)

            and    OM perpendicular to  CD

           =CM=1/2 CD ..... (ii)

           But, AB = CD

            1/2AB=1/2CD

            AL =CM ..... (iii)          [Using equations (i) and (ii)]

           Now, in right triangles OAL and OCM, we have

           OA= OC                    [Equal to radius of the circle]

           AL= CM                   [From equation (iii)]

          &  ∠ALO =∠CMO      [Each equal to 90°]

         So, by RHS criterion of congruence,

        ⊿OAL= ⊿OCM

        OL= OM

      Hence, equal chords of a circle are equidistant from the centre.

    Theorem 4 : Equal chords of a circle subtend equal angles at the centre.

Given  A circle C(O, r) and its two equal chords AB and CD.

To Prove   ∠AOB =∠COD

Proof         In triangle AOB and COD, we have

                  AB= CD                [Given]

                OA=OC                  [Each equal to r]

                OB= OD                 [Each equal to r]

              So, by SSS-criterion of congruence, we have

               ⊿AOB = ⊿COD

              =∠AOB=∠COD

Theorem 5 : If the angles subtended by two chords of a circle at the centre are equal, the chords are equal.

Given    Two chords AB and CD of a circle C(O, r) such that ∠AOB = ∠COD

To Prove       AB= CD

Proof         In triangle AOB and COD, we have

                 OA= OC          [Each equal to r]

                ∠AOB= ∠COD                  [Given]

                 OB= OD                         [Each equal to r]

               So, by SAS-criterion of congruence,

               ⊿AOB = ⊿COD

                AB=CD

Theorem 6 : The angle subtended by an arc of a circle at the centre is double the angle subtended by it at

any point on the remaining part of the circle.

Given      An arc PQ of a circle C(O, r) and a point R on the remaining part of the circle i.e. arc QP.

To Prove   ∠POQ = 2∠PRQ

Construction       Join RO and produce it to a point M outside the circle.

Proof   We shall consider the following three different cases:

             Case-I When arc PQ is minor arc

            We know that an exterior angle of a triangle is equal

            to the sum of the interior opposite angles.

            In⊿ POR, ∠POM is the exterior angle

           :.∠ POM=∠ OPR + ∠ORP                                 [ OP=OR=r :. ∠O PR= ∠ ORP]

          ∠POM=2 ∠ORP .. .. (i)

           In ⊿QOR,  ∠QOM is the exterior angle

          ∠QOM=  ∠OQR+  ∠ORQ

         =∠QOM= ∠ORQ+∠ORQ      [ OQ=OP=r :.∠OPQ=∠OQP

          ∠QOM=2∠ORQ   ..... (ii)

Adding cquations (i) and (ii), we get

 ∠POM+  ∠QOM=2 ∠ORP+2 ∠ORQ

 ∠POM +  ∠QOM=2( ∠ORP +  ∠ORQ)

 ∠POQ =2 ∠PRQ