Two circles with centres A and …

Since tangent at a point to a circle is perpendicular to the radius through the point of contact. Therefore, $\angle A C B$ = 90°
In $\Delta A C B$, we have
  AB² = AC² + BC² [By Pythagoras Theorem]
$\Rightarrow$ AB² = 3² + 4² = 9 + 16 = 25
AB = 5 cm
Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.
$\therefore \quad A P \perp C D$
and $C P = P D$
Let AP = x. Then, BP = AB - AP = 5 - x cm (AB = 5cm)
Further, let CP = DP = y cm. 
In ∆ APC and ∆ BPC applying Pythagoras theorem, we have

    AC² = A P² + PC², BC² = PB² + PC²
$\Rightarrow \quad$ 3² = x² + y² ......(1)

and, 4² = (5 - x)² + y²......(2)

Subtracting equation (1) from equation (2) , we get :-
$\Rightarrow$ 4² - 3² = {(5 - x)² + y²} - {x² + y²} 
$\Rightarrow$ 7 = 25 - 10x
$\Rightarrow$ 10x = 18 $\Rightarrow$ x = 1.8 cm

Now, substituting x = 1.8 in equation (1) , we get :-

       3² = (1.8)² + y²

 $\Rightarrow$ y = $\sqrt { 9 - ( 1.8 ) ^ { 2 } } = \sqrt { 5.76 } = 2.4 \mathrm { cm }$

Hence, common chord  CD = 2CP = 2y = 4.8 cm.   Ans.