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.

23.58 units

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.