First of all there is actually no length of a tangent, it can be extended to any length. and if you are asking about the length of the line which touches both circle, which is apart of the line which joins the center of two circles, then the correct answer is
"D - (R-r)"
where 'D' is the distance between centers of the two circles. 'R' and 'r' are the radii of two circles

Drawing the Circles

To solve this problem, let's first draw the two circles with their centers and radii. Let's assume that the center of the first circle is called C1, and the center of the second circle is called C2. The radii of the circles are r and R, respectively. The distance between the centers of the circles is represented by D.

Locating the Tangent

Now, let's draw a line segment connecting the centers of the circles. This line segment will have a length equal to D. Next, draw a perpendicular bisector to this line segment. This perpendicular bisector will intersect the line joining the two centers at a point, which we will call P.

Determining the Length of the Common Tangent

The point P is equidistant from the centers of both circles, so it lies on the perpendicular bisector. Since P is equidistant from the centers of the circles, the lengths of PC1 and PC2 are both equal to the average of the radii, which is (r + R)/2.

Now, let's draw a line segment from P to the point where the first circle touches the common tangent. Let's call this point T1. Similarly, draw a line segment from P to the point where the second circle touches the common tangent. Let's call this point T2.

Applying the Pythagorean Theorem

Since the line segments PT1 and PT2 are perpendicular to the common tangent, we can apply the Pythagorean theorem to find the lengths of these line segments. The square of PT1 is equal to the square of PC1 minus the square of CT1. Similarly, the square of PT2 is equal to the square of PC2 minus the square of CT2.

Using the Pythagorean theorem, we can write the following equations:

PT1^2 = PC1^2 - CT1^2
PT2^2 = PC2^2 - CT2^2

Since PT1 and PT2 are equal in length (as they are both tangents from the same point), we can write:

PC1^2 - CT1^2 = PC2^2 - CT2^2

Substituting the values of PC1 and PC2, we get:

((r + R)/2)^2 - CT1^2 = ((r + R)/2)^2 - CT2^2

Simplifying the equation, we find:

CT1^2 = CT2^2

Taking the square root of both sides, we get:

CT1 = CT2

Therefore, the lengths of the line segments CT1 and CT2 are equal, which means that the length of the common tangent touching both circles is CT1 or CT2.