Ex 1:

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.

Sol:

Given AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
To prove:AP = PD and PB = CP.
Construction: Draw OM perpendicular to AB and ON perpendicular to CD. Join OP.

AM = MB = AB (Perpendicular bisecting the chord)
CN = ND = CD (Perpendicular bisecting the chord)
AM = ND and MB = CN (As AB = CD)
In triangle OMP and ONP, we have,
OM = MN (Equal chords are equidistant from the centre)
∠OMP = ∠ONP (90°)
OP is common. Thus triangle OMP and ONP are congruent (RHS).
MP = PN (cpct)
So, AM + MP = ND + PN
or, AP = PD .......... (i)
As MB = CN and MP = PN,
MB - MP = CN - PN
PB = CP ....... (ii)
Ex 2:

If two equal chords of a circle intersect within a circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Sol:
Let AB and CD are the two equal chords of a circle having centre 'O'.
Again let AB and CD intersect each other at a point M.
Now, draw OP perpendicular to AB and OQ perpendicular to CD.
From the figure,
In ΔOPM and ΔOQM,
OP = OQ {equal chords are equally distant from the center}
∠OPM = ∠OQM
OM = OM {common}
By SAS congruence criterion,
ΔOPM ≅ ΔOQM
So, ∠OMA = ∠OMD
or ∠OMP = ∠OMQ {by CPCT}
Ex 3:

If a line intersects two concentric circles (circles with the same centre) with centre O at P, Q, R and S, prove that PQ = RS

Sol:

Let us draw a perpendicular OM is on line PS
we know that perpendicular drawn from the centre of the circle bisects the chord
QM = MR ..... (i)
PM = MS ....... (ii)
On subtracting (ii) from (i), we get
PM – QM = MS – MR
PQ = RS

Ex 4:

Two circles of radii 17 cm and 25 cm intersect at two points and the distance between their centres is 28 cm. Find the length of the common chord.

Sol:
Radius of the circle with centre P is AP = 17 cm
The radius of the circle with centre O is OA = 25 cm
The distance between their centres is PO = 28 cm
Let PC = x cm
In ΔACP,
AP2 = AC2 + PC2
(17)2 = AC2 + x2
AC2 = (17)2 – x2 ..... (i)
In ΔACO,
AO2 = AC2 + CO2
(25)2 = AC2 + (28 – x)2
AC2 = (25)2 – (28 – x)
= 625 – (784 + x2 – 56x)
= – 159 – x2 + 56x ..... (ii)
From (i) and (ii)
(17)2 – x2 = – 159 – x2 + 56x
56x = 448
x = 8
From eq (i)
AC2 = 172 – 82
AC = 15
From the figure, OP ⊥ AB
AC = CB
AB = AC + CB
= 2AC = 30
Ex 5:

Three girls Roja, Shanthi and Madhu are playing a game by standing on a circle of radius 4 m drawn in a park. Roja throws a ball to Shanthi, Shanthi to Madhu, Madhu to Roja. If the distance between Roja and Shanthi and between Shanthi and Madhu is 4 m each, what is the distance Roja and Madhu?

Sol:

Draw perpendiculars OA and OB on RS and SM respectively.
AR = AS = 4/2 = 2 m
OR = OS = OM = 4 m (Radii of the circle)
In ΔOAR,
OA2 + AR2 = OR2
OA2 + 22 = 42
OA2 = 16 – 4 = 12
OA = 2√3 = 3.4
ORSM will be a kite (OR = OM and RS = SM).
We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
∠RCS will be of 90° and RC = CM
Area of ΔORS = × OA × RS
× RC × OS = × 2√3 × 4
RC = 2√3
RC = 3.4

Ex 6:

A circular park of radius 30 m is situated in a colony. Three boys Arun, Sham and Dev are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Sol:

In this given that AS = SD = DA
Therefore, ΔASD is an equilateral triangle, OA = 30 m
The median of equilateral triangle pass through the circumcentre (O) of the equilateral triangle ASD.
We also know that medians intersect each other in the ratio 2 : 1.
As AB is the median of equilateral triangle ASD,
We can write
AB = OA + OB = (30 + 15) = 45 m
In ΔABD
AD2 = AB2 + BD2
AD2 = (45)2 + (AD/2)2
AD2 = 2025 + (1/4)AD2
(3/4)AD2 = 2025
AD = 30√3

Ex 7:

Of any two chords of a circle, show that the one which is large is nearer to the centre.

Sol:

Given AB and CD are two chords of a circle C(O, r) such that AB > CD.
Also, OE ⊥ AB and OF ⊥ CD
To prove OE < OF
Construction join OA and OC
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AE = AB
and CF = CD ------- (i)
Also, OA = OC = r ----- (ii)
And, AB > CD
⇒ CD < AB
⇒ CD < AB ⇒ CF < AE ------ (iii) [from (i)]
Now, from the right angles triangles Δ OEA and Δ OFC, we have:
OA2 = OE2 + AE2 and
OC2 = OF2 + CF2
⇒ OE2 + AE2 = OF2 + CF2 [∵ OA = OC ⇒ OA2 = OC2]
⇒ OE2 + AE2 < OF2 + AE2 [using (iii)]
⇒ OE2 < OF2
⇒ OE < OF