Question

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 chrod.​

Answer 1

Step-by-step explanation:

Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'

Hence OA = OB = 5 cm

O'A = O'B = 3 cm

OO' is the perpendicular bisector of chord AB.

Therefore, AC = BC

Given OO' = 4 cm

Let OC = x

Hence O'C = 4 − x

In right angled ΔOAC, by Pythagoras theorem

OA2 = OC2 + AC2

⇒ 52 = x2 + AC2

⇒ AC2 = 25 − x2 à (1)

In right angled ΔO'AC, by Pythagoras theorem

O'A2 = AC2 + O'C2

⇒ 32 = AC2 + (4 – x)2

⇒ 9 = AC2 + 16 + x2 − 8x

⇒ AC2 = 8x − x2 − 7 à (2)

From (1) and (2), we get

25 − x2 = 8x − x2 − 7

8x = 32

Therefore, x = 4

Hence the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.

AC2 = 25 − x2  

= 25 − 42  

= 25 − 16 = 9

Therefore, AC = 3 m

Length of the common chord, AB = 2AC = 6 m

Answer 2

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Let bigger circle center be 0 and smaller I,e 3cm radius be 0'

Now from the figures OA=5cm =OB

O'A=3cm=O'B

The line OO' is a perpendicular bisector of AB.

Therefore AC=CB

And it's given OO'=4cm

Let OC be X.

So, the value of O'C =4-x

In the triangle OAC

OA^2=AC^2+OC^2

5^2=AC^2+X^2

25-X^2=AC^2...(1)

Again in triangle O'AC

O'A^2=AC^2+O'C^2

3^2=AC^2+(4-x)^2. (since O'C=4-x)

9=AC^2+16-8x+X^2

AC^2=-X^2+8X-7..(2)

Comparing equation (1)and (2) we got

that they are equal.

25-X^2=-X^2+8x-7

25=8x-7 (x^2 get cancelled)

=>8X=25+7

=>X=32/8

=>X=4

So, replacing this value in eq(1)

AC^2=25-X^2

AC^2=25-4^2

AC^2=25-16

AC^2=9

AC=3

$\underline\bold{So, length \:of \:common \:cord\: is\: 2×AC=6cm\: :}$

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