pl expalin me dis question properly of ex 10.4 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.

Given, two circles of radii 5 cm and 3 cm intersecting at two points and the distance between their centers is 4 cm and we have to find out the length of the common chord.

In order to solve it firstly we draw a figure as shown below. Here, O and O' is the center of the circles having radii 5 cm and 3 cm respectively.

Now, it is given that distance between the center of the circles = OO' = 4 cm

OA = OB = 5 cm  [radii of the larger circle]

 O'A = O'B = 3 cm [radii of the smaller circle]

We know that, if two circles intersect in two points then the line through the center is perpendicular bisector of the common chord.

⇒ OO' will be the perpendicular bisector of chord AB.

⇒ AC = CB

It is given that, OO' = 4 cm

Consider OC = x. Therefore, O'C will be x -4.

In ΔOAC by pythagoras theorem, we have

OA² = AC² + OC²

⇒ 5² = AC² + x²

⇒ 25 - x² = AC² ... (1)

Again, in ΔO'AC by pythagoras theorem, we have

O'A² = AC² + O'C²

⇒ 3² = AC² + (x - 4)²

⇒ 9 = AC² +  x² + 16 - 8x

⇒ AC² = - x² - 7 + 8x ... (2)

On comparing the result of equations (1) and (2), we get

25 - x² = - x² - 7 + 8x

⇒ 25 + 7 = 8x

⇒8x = 32

x = 4

Therefore, the common chord will pass through the center of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

AC² = 25 - x² = 25 - 4² = 25 - 16 = 9

⇒ AC = 3 m

Length of the common chord AB = 2 AC = (2 × 3) m = 6 m

Hope you get it!!