Question

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

Answer 1

let O and O' two circle .
which intersect in A and B .
so, AB is common chord .

we know,
chord is perpendicularly bisected by line joining of center to its .
let line meet at T

now,
∆ OAT is right angle ∆
so,
length of OT =√{(5^2 -(x/2)^2 }

where x is length of chord

again ,
for ∆ O' AT

length of O'T =√{(3)^2 -(x/2)^2

but here ,
length of OT + length of O'T =distance between centre of circles

√(25 - x^2/4) +√(9 -x^2/4 ) =4
let
x^2/4 =r
√(25-t) +√(9-t) =4

if we put t = 9
then,
√(25 -9) +√(9-9) = √16 +0 =4
LHS = RHS

so,

t =9
x^2/4 =9
x^2 =36

x=6 cm

so,
length of chord = 6 cm

Answer 2

Let AB be a line segment of 4cm
A is the centre of the cirlce with radius 5 cm
B is the centre of circle with radium 3 cm
CD is the common chord.

But the chord passes through the centre of CircleA
∴ It is the diameter of circle A
∴ 2(3cm)= 6cm
∴CD=6cm

OR
∠ACB is a right angle because any line from the centre to a chord is ⊥ to the chord and bisects it.
By Pythogora's theorem,
Hypotenuse² = side1² + side2²
∴ in ΔACB
CA²=AB²+CB²
5²=4²+CB²
25=16+CB²
9=CB²
3cm=CB

∵CB = BD
∴CD = 2CB
∴CD = 2(3) cm
∴CD = 6cm