# 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.

Let O and O' be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

Given: OA = 5cm, O'A = 3cm and OO' = 4cm

In $$\triangle{OO'A}$$, we have,
$${AO'}^2 + {OO'}^2 = {3}^2 + {4}^2 = 9 + 16 = 25$$
$$\Rightarrow$$ $${OA}^2 = {5}^2 = {AO'}^2 + {OO'}^2$$

Therefore, OO'A is a right angled traingle and right angled at O'

Now, Area of $$\triangle{OO'A}$$ = $$\frac{1}{2}$$ × O'A × OO'
= $$\frac{1}{2}$$ × 3 × 4

Area of $$\triangle{OO'A}$$ = 6 sq.cm ...(i)
Also, Area of $$\triangle{OO'A}$$ = $$\frac{1}{2}$$ × OO' × AM
= $$\frac{1}{2}$$ × 4 × AM
Area of $$\triangle{OO'A}$$ = 2 × AM ...(ii)

From eq. (i) and (ii), we get,
2 × AM = 6

Therefore, AM = 3 cm.

Since, when two circles intersect at two points, then their centres lie on the perpendicular bisector of the common chord.

Thus, AB= 2 × AM = 2 × 3 = 6cm.

Therefore, the length of the common chord is 6cm.