# PROBLEMS BASED ON CIRCLES EXAMPLES

## About "Problems Based on Circles Examples"

Problems Based on Circles Examples :

Here we are going to see some example problems based on circles.

## Problems Based on Circles Examples - Practice questions

Question 1 :

Th e diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.

Solution :

Diameter  =  52 cm

Radius of circle  =  26 cm

In triangle BOC,

OC2  =  OB2 + BC2

262  =  OB2 + 102

676 - 100  =  OB2

OB  =  576

OB  =  24

Question 2 :

The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle

Solution :

In triangle BOC,

OC2  =  OB2 + BC2

OC2  =  82 + 152

OC2  =  64 +  225

OC  =  √289

OC  =  17

Question 3 :

Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 42 cm and also find <OAC and <OCA.

Solution :

AC =  AO2 + CO2

AC =  (4√2)2 + (4√2)2

AC =  16(2) + 16(2)

AC =  64

AC  =  √64

AC  =  8 cm

In triangle AOC,

<COA  =  90

OC  =  OA

Equal sides will form a equal angles.

<OCA  =  <OBC

Question 4 :

A chord is 12 cm away from the centre of the circle of radius 15 cm. Find the length of the chord.

Solution :

Let BC  =  x

OC2  =  OB2 + BC2

152  =  122 + x2

225  -  144  =  x2

x  =  √81

x  =  9 cm

AC  =  2BC  =  2(9)  =  18 cm

Question 5 :

In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?

Solution :

In triangle OPB,

OB2  =  OP2 + PB2

OA  =  OB  =  OC  =  OD  =  10 (Radius)

102  =  OP2 + 82

OP2  =  100 - 64

OP2  =  36

OP  =  √36

OP  =  6 cm

In triangle OQD,

OD2  =  OQ2 + QD2

102  =  OQ2 + 62

OQ2  =  100 - 36

OQ2  =  64

OQ  =  √64

OQ  =  8 cm

Distance between two chords  PQ  =  OQ - OP

=  8 - 6

=  2 cm

Question 6 :

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution :

OP  =  3 cm, PS  =  5 cm and OS  =  4 cm

Let RS be x

OR  =  4 - x

In triangle POR,

OP =  OR2 + PR2

3 =  x2 + PR2

PR2  =  9 - x  ------(1)

In triangle PRS,

PS2  =  PR2 + RS2

52  =  PR2 + (4 - x)2

PR2  =  25 - (4 - x)2

PR2  =  25 - (16 - 8x + x2)

PR2  =  25 - 16 + 8x - x2

PR2  =  9 + 8x - x ----(2)

(1)  =  (2)

9 - x2  =  9 + 8x - x

8x  =  0

x  =  0

By applying the value of x in (1), we get

PR2  =  9 - 0

PR2  =  9  ==>  PR  =  3

PQ  =  2PR

PQ  =  2(3)

PQ  =  6 cm

After having gone through the stuff given above, we hope that the students would have understood, "Problems Based on Circles Examples"

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