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Find the length of direct common tangent
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A. 98
B. 99
C.10
D. 101

Answer
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Hint: This question can be easily answered by constructing a tangent which makes a right angle with the radius of the both circles and using this given information you will get a better approach to solution to the given problem.

Complete step-by-step answer:
To find the solution of the given question first construct and common tangent to both the circles with center C1 and C2 respectively and name the points that are unknown in the given diagram

So after observing the above diagram we have C1C is equal to 4 cm, C2D is equal to 3 cm and CD is equal to 3 cm
Since after observing the above diagram we know that the line C1C and C2D are also perpendicular to the radius of the circles
Constructing a line EB equal to the length of C1C2 where E is the point on line AC1
So to find the length of AB we have to find the length of EB and AE
So since we know that EB=C1C2
So EB = C1C + CD + C2D (equation 1)
After substituting the given values in equation 1 we get
EB = 4 + 3 + 3
  EB = 10 cm
And since AC1 is equal to AE + EC1 here EC1 is equal to BC2 so EC1 = 3 cm
Now substituting the values in the equation AC1 = AE + EC1
4 = AE + 3
  AE = 4 – 3
AE = 1cm
In triangle AEB by Pythagoras theorem
 (AB)2=(AE)2+(EB)2
Substituting the given values in the above formula
 (AB)2=(1)2+(10)2
  (AB)2=1+100
  AB=101 cm
So the length of the common tangent is 101 cm
So, the correct answer is “Option D”.

Note: The concept of tangent played a main role to solve the above problem which can be explained as the line which intersect circle at one point that lie on the circumference of the circle here at the point where tangent intersect the circle, radius of the circle becomes perpendicular to the tangent. There are some basic properties that are shown by the tangent such as: the line of tangent never passes through the circle and the tangent drawn from the external points of the circle are equal to each other.
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