1100 -cycle) ě Yadit 5 Crn and 3cmintersect at two beirls and the

1.	1100 -cycle) ě Yadit 5 Crn and 3cmintersect at two beirls and the diztence
Answer» Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively. OA = OB = 5 cm O'A = O'B = 3 cm OO' will be the perpendicular bisector of chord AB. ∴ AC = CB It is given that, OO' = 4 cm Let OC bex. Therefore, O'C will be 4 −x. In ΔOAC, OA2= AC2+ OC2 ⇒ 52= AC2+x2 ⇒ 25 −x2= AC2... (1) In ΔO'AC, O'A2= AC2+ O'C2 ⇒ 32= AC2+ (4 −x)2 ⇒ 9 = AC2+ 16 +x2− 8x ⇒ AC2= −x2− 7 + 8x... (2) From equations (1) and (2), we obtain 25 −x2= −x2− 7 + 8x 8x= 32 x= 4 Therefore, the common chord will pass through the centre 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

1100 -cycle) ě Yadit 5 Crn and 3cmintersect at two beirls and the diztence

Answer»

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

∴ AC = CB

It is given that, OO' = 4 cm

Let OC bex. Therefore, O'C will be 4 −x.

In ΔOAC,

OA2= AC2+ OC2

⇒ 52= AC2+x2

⇒ 25 −x2= AC2... (1)

In ΔO'AC,

O'A2= AC2+ O'C2

⇒ 32= AC2+ (4 −x)2

⇒ 9 = AC2+ 16 +x2− 8x

⇒ AC2= −x2− 7 + 8x... (2)

From equations (1) and (2), we obtain

25 −x2= −x2− 7 + 8x

8x= 32

x= 4

Therefore, the common chord will pass through the centre 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