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**Hint:**According to the question we have to find the length of direct common tangent to the two circles will be when the circles of radii 2 and 3 cm touch each other externally. So, first of all we have to draw a diagram from the given circles and the direct common tangent which is as given below:

In the figure give above, radius of smaller circle $AP = 2cm$ and the bigger circle is $BQ = 3cm$ and PQ is the direct common tangent of the two circles and as we can see from the diagram line AP is parallel to the line BQ and as we know that line AP is perpendicular to the common tangent PQ and line BQ is perpendicular to the tangent PQ hence, $\angle APQ = \angle RQP$ which are equal to the right angles.

Now from the above diagram we will obtain the rectangle APQR.

Now, we can see from the diagram that AB is the sum of the radius of both smaller and bigger circles and after that we have to find the distance or the length of line BR.

Now, we have to consider the triangle ARB which is right angled at B hence, on finding the length of AR we can obtain the length of PQ which is the direct common tangent.

**Complete step-by-step solution:**

Step 1: First of all we have to construct the diagram for to circles having radii 2 and 3 cm and a common tangent as explained in the solution hint which is drawn below:

Step 2: Now, as explained in the solution hint that line AP is parallel to line BQ and angle P is equal to the angle Q. Hence,

$ \Rightarrow AP\left\| {BQ..................(1)} \right.$ and,

$ \Rightarrow \angle APQ = \angle RQP = {90^0}.................(2)$

Step 3: Now, as explained in the solution hint that line AR is perpendicular to line BQ and line PQ is perpendicular to line BQ. Hence,

$ \Rightarrow PQ\left\| {AR...................(3)} \right.$

Which means line PQ is parallel to the line AR.

Step 4: Now, from the equation (1), (2), and (3) it's evident that APQR is a rectangle as mentioned in the solution hint. Hence,

$ \Rightarrow PQ = AR$ and $AP = RQ......................(4)$

Step 5: Now, as we can see in the diagram AB is the sum of the radii of the two circles and since, the distance between the centres of two circles, which touches externally, is the sum of the radii of the two circles. Hence,

$

\Rightarrow AB = (2 + 3)cm \\

\Rightarrow AB = 5cm

$

Now, we can also obtain the length of line BR as below with the help of equation (4),

$

\Rightarrow BR = BQ - RQ \\

\Rightarrow BR = (3 - 2)cm \\

\Rightarrow BR = 1cm

$

Step 5: Now, we have to consider the triangle ARB which is a right angled triangle as explained in the solution hint. Hence,

$ \Rightarrow AR = \sqrt {A{B^2} - B{R^2}} $

On substituting all the values,

$

\Rightarrow AR = \sqrt {{5^2} - {1^2}} \\

\Rightarrow AR = \sqrt {25 - 1} \\

\Rightarrow AR = \sqrt {24} \\

\Rightarrow AR = 2\sqrt 6 cm

$

Step 6: Now, from the equation (4) we can obtain the value of line PQ which is a direct common tangent. Hence,

$ \Rightarrow PQ = 2\sqrt 6 cm$

Final solution: Hence, we have obtained the length of direct common tangent to the two circles when the circles of radii 2 and 3 cm touch each other externally is $2\sqrt 6 cm$.

**Therefore option (A) is correct.**

**Note:**Common tangent is a tangent to a circle or the circle is defined as a line passing through exactly one point of the circle and it is perpendicular to the line which is passing through the centre of the circle.

A line that is a tangent to more than one circle is considered as a common tangent for both of the given circles.

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