AB is a common tangent to both the circles in the given figure. Find the distance (correct to two decimal places) between the centres of the two circles.
Answer & explanation
Correct answer: option 2
Concept Used
1.Two triangles are said to be similar if they have the same ratio of corresponding sides and equal pairs of corresponding angles.
2. Pythagoras theorem
Calculation
By using pythagoras theorem,
CE = √(\( { CA}^{ 2} \) + \( { AE}^{ 2} \))
CE = √(\( { 5}^{ 2} \) + \( { 8}^{ 2} \))
CE = \(\sqrt { 89}\) = 9.43 units
and,
ED = √(\( { BE}^{ 2} \) + \( { BD}^{ 2} \))
ED = √(\( { 12}^{ 2} \) + \( { x}^{ 2} \))
From the figure,
\(\Delta \)ECA is similar to \(\Delta \)EDB
Using the property of the similarity of a triangle,
\(\frac{BD}{AC}\) = \(\frac{BE}{AE}\)
\(\frac{x}{5}\) = \(\frac{12}{8}\)
x = \(\frac{60}{8}\) = 7.5 units
Now, substituting the value of x in equation (1) we get,
ED = √(\( { 12}^{ 2} \) + \( { 7.5}^{ 2} \))
ED = \(\sqrt { 200.25}\) = 14.15 units
Distance between the centers of the two circles = CE + ED
= 9.43 + 14.15
= 23.58 units
Therefore, Distance between the centers of the two circles is 23.58 units.