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

The correct option is C: $6cm$

Let $OP=5cm$ be the radius of the circle $C_1$ and let $PT=3cm$ be the radius of the circle $C_2$

Let both circle intersect at points $P$ and $Q$ such that $PQ$ is the common chord of the circles.

Let $OT$ intersect the common chord $PQ$ at point $S$.

Now consider $\Delta OPT$ and $\Delta OQT$,

$\Rightarrow OP=OQ$ [Radii of the same circle]

$\Rightarrow PT=TQ$ [Radii of the same circle]

$\Rightarrow OT=OT$ [Common side]

Therefore, $\Delta OPT\cong \Delta OQT$ [by SSS congruence rule]

$\Rightarrow \angle POT=\angle QOT$ [By CPCT] ...(i)

Now, consider $\Delta POS$ and $\Delta QOS$

$\Rightarrow OP=OQ$ [Radii of the same circle]

$\Rightarrow \angle POT=\angle QOT$ [From (i)]

$\Rightarrow OS=OS$ [Common side]

Therefore, $\Delta POS\cong \Delta QOS$ [by SAS congruence rule]

$\Rightarrow PS=SQ$ i.e. radii of both the circles bisect common chord $PQ$.

Hence, $\angle PSO=\angle PST={90}^{\circ }$ [If radius bisects the chord then it must be perpendicular also.]

Therefore, both $\Delta PSO$ and $\Delta PST$ are right angled triangle.

Now let, $TS=x$ such that $OS=4-x$ [As $OT=4$ cm]

Consider right $\Delta PSO$,

$\Rightarrow O{P}^2=P{S}^2+O{S}^2$

$\Rightarrow 5^2=P{S}^2+(4-x{)}^2$

$\Rightarrow 25=P{S}^2+16+x^2-8x$

$\Rightarrow P{S}^2=-x^2+8x+9$ ...(ii)

Consider right $\Delta PST$,

$\Rightarrow P{T}^2=T{S}^2+P{S}^2$

$\Rightarrow 3^2=x^2+P{S}^2$

$\Rightarrow P{S}^2=9-x^2$...(iii)

From (ii) and (iii), we have

$\Rightarrow -x^2+8x+9=9-x^2$

$\Rightarrow 8x=0$

$\Rightarrow x=0$

This means that point $S$ and $T$ coincide with each other such that $OS=OT=4$ cm

Therefore, from (iii), we have

$\Rightarrow P{S}^2=9^2$

$\Rightarrow PS=3cm$

Hence, $PQ=2\times PS=6cm$