**Problems Based on Circles Examples :**

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

**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,

OC^{2} = OB^{2} + BC^{2}

26^{2} = OB^{2} + 10^{2}

676 - 100 = OB^{2}

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,

OC^{2} = OB^{2} + BC^{2}

OC^{2} = 8^{2} + 15^{2}

OC^{2} = 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 4√2 cm and also find <OAC and <OCA.

**Solution :**

AC^{2 } = AO^{2} + CO^{2}

AC^{2 } = (4√2)^{2} + (4√2)^{2}

AC^{2 } = 16(2) + 16(2)

AC^{2 } = 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

OC^{2} = OB^{2} + BC^{2}

15^{2} = 12^{2} + x^{2}

225 - 144 = x^{2}

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,

OB^{2} = OP^{2} + PB^{2}

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

10^{2} = OP^{2} + 8^{2}

OP^{2} = 100 - 64

OP^{2} = 36

OP = √36

OP = 6 cm

In triangle OQD,

OD^{2} = OQ^{2} + QD^{2}

10^{2} = OQ^{2} + 6^{2}

OQ^{2} = 100 - 36

OQ^{2} = 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^{2 } = OR^{2} + PR^{2}

3^{2 } = x^{2} + PR^{2}

PR^{2} = 9 - x^{2 } ------(1)

In triangle PRS,

PS^{2} = PR^{2} + RS^{2}

5^{2} = PR^{2} + (4 - x)^{2}

PR^{2} = 25 - (4 - x)^{2}

PR^{2} = 25 - (16 - 8x + x^{2})

PR^{2} = 25 - 16 + 8x - x^{2}

PR^{2} = 9 + 8x - x^{2 } ----(2)

(1) = (2)

9 - x^{2 }= 9 + 8x - x^{2 }

8x = 0

x = 0

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

PR^{2} = 9 - 0

PR^{2} = 9 ==> PR = 3

PQ = 2PR

PQ = 2(3)

PQ = 6 cm

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