Question

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

Answer 1

Answer:

The perpendicular bisector of the common chord passes through the centers of both circles.

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

Given that the circles intersect at two points, so we can draw the above figure. Let AB be the common chord. Let O and O’ be the centers of the circles, respectively.

O’A = 5 cm, OA = 3 cm

OO’ = 4 cm [Given distance between the centres is 4cm]

Since the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of the smaller circle lies inside, the bigger circle itself.

OO’ is the perpendicular bisector of AB.

So, OA = OB = 3 cm

AB = 3 cm + 3 cm = 6 cm [Since, O is the mid point of AB]

The length of the common chord is 6 cm.

Answer 2

$AP^2 =AR^2 + PR^2 \\ AR^2 = 5^2 - x^2(1)$

In △ARQ,

$AQ^2 = AR^2 + QR^2</p><p>AR^2 = 3^2 - (4-x)^2(2)</p><p>$

$∴ {5}^{2} - {x}^{2} = {3}^{2} - (4 - x)\: from \: (1) \: and \: (2) \\ 25 - {x}^{2} = 9 - (16 - 8x + {x}^{2} ) \\ 25 - {x}^{2} = - 7 + 8x - {x }^{2} \\ \: \: \: \: \: \: 32 = 8x \\ ∴x = 4$

Substitute in eq (1) we get

$∴AR^2=25-16-9 \\ ∴AR =3 cm \\ ∴ AB = 2 \times AR = 2 \times 3</p><p>\\ ∴</p><p>AB = 6cm</p><p></p><p>$