Theory:

Result \(4\):
If  two circles touch externally, the distance between their centres is equal to the sum of their radii.
Explanation:
 
r4.png
 
If  two circles touch externally at a point \(B\), then the distance \(OA\) is equal to the sum of the radii \(OB\) and \(AB\).
 
\(\Rightarrow\) \(OA\) \(=\) \(OB\) \(+\) \(AB\)
Proof for the result:
Let the two circles with centres \(O\) and \(A\) intersect each other externally at the point \(B\).
 
Let the radius \(OB\) \(=\) \(r_{1}\) and \(AB\) \(=\) \(r_{2}\) and \(r_{1}\) \(>\) \(r_{2}\).
 
Let the distance between the centres be \(d\).
 
\(\Rightarrow\) \(OA\) \(=\) \(d\)
 
From the figure, we observe that \(OA\) \(=\) \(OB\) \(+\) \(AB\).
 
\(\Rightarrow\) \(d\) \(=\) \(r_{1}\) \(+\) \(r_{2}\)
Example:
Two circle with radii \(4\) \(cm\) and \(5\) \(cm\) intersect at a point \(O\) externally. If so, find the distance between their centres.
 
Solution:
 
By the result, we know that:
 
Distance between the centres \(=\) Sum of the radii.
 
Thus, the distance between the centres \(=\) \(4\) \(cm\) \(+\) \(5\) \(cm\)
 
\(=\) \(9\) \(cm\)