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

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2 cm

4 cm

5 cm

6 cm

answer is D.

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Detailed Solution

Let the diagram of the given situation be,
Question Image

Given that,

O P = 5 c m

and

Y P = 3 c m

.
In

Δ O P Y

and

Δ O Q Y

, we have
The radius of circle 1 is

O P = O Q

, the radius of circle 2 is

P Y = Q Y

, and the common side is

O Y = O Y

.
By SSS congruent rule, we have

O P Y ≅ O Q Y

. By C.P.C.T, we get

∠ P O Y = ∠ Q O Y

… (1)
In

Δ P O X

and

Δ Q O X

, we have
The radius of circle 1 is

O P = O Q

, by (1), we have

∠ P O X = ∠ Q O X

, and the common side is

O X = O X

.
By SSS congruent rule, we have

Δ P O X ≅ Δ Q O X

. By C.P.C.T, we get

∠ P X O = ∠ Q X O

… (2) and

P X = Q X

… (3)
As PQ is a straight line, then

∠ P X O + ∠ Q X O = 180 ° ∠ P X O + ∠ P X O = 180 ° 2 ∠ P X O = 180 ° ∠ P X O = 90 ° ∠ P X O = ∠ Q X O = 90 °

Thus,

△ P X O

and

△ P X Y

is a right-angled triangle.
Let

O X = x

and

X Y = 4 − x

. In

△ P X O

and

△ P X Y

, by using the Pythagoras theorem, we get

O P 2 = O X 2 + P X 2 52 = x 2 + P X 2 52 − x 2 = P X 2 P Y 2 = X Y 2 + P X 2 32 = (4 − x) 2 + P X 2 32 − (4 − x) 2 = P X 2

Equating the values of

P X 2

, we have

32 − (4 − x) 2 = 52 − x 2 9 − (16 + x 2 − 8 x) = 25 − x 2 8 x = 32 x = 4

To find the value of PX, substitute the value of x in the equation and we have

P X 2 = 52 − 42 = 25 − 16 P X = 9 P X = 3 c m

The required length of common chord is given by,

= P X + Q X = P X + P X = 2 P X = 2 (3) = 6 c m

Hence, the correct option is 4.

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