Question : 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.
Option 1: 18.98 units
Option 2: 23.58 units
Option 3: 26.59 units
Option 4: 21.62 units
Correct Answer: 23.58 units
Solution :
Given,
The radius of the smaller circle = 5 units
Let the radius of the larger circle be $x$ units.
Common tangent(AB) to both circles of length = 20 units
We know that two triangles are said to be similar if they have the same ratio of corresponding sides and equal pairs of corresponding angles.
Using Pythagoras theorem,
⇒ $CE = \sqrt{(CA)^2 + (AE)^2}$
⇒ $CE = \sqrt{(5)^2 + (8)^2}$
⇒ $CE = \sqrt{89}$
⇒ $CE = 9.43$ units
Also,
$ED = \sqrt{(BE)^2 + (BD)^2}$
⇒ $ED = \sqrt{(12)^2 + (x)^2}$ --------(1)
From the figure,
$△ECA$ is similar to $△EDB$.
Using the property of the similarity of a triangle,
⇒ $\frac{BD}{AC}=\frac{BE}{AE}$
⇒ $\frac{x}{5}=\frac{12}{8}$
⇒ $x = \frac{60}{8}$
⇒ $x=7.5$ units
Now, substituting the value of $x$ in equation (1) we get,
⇒ $ED = \sqrt{(12)^2 + (7.5)^2}$
⇒ $ED = \sqrt{200.25}$
⇒ $ED =14.15$ units
$\therefore$ Distance between the centres of the two circles
= CE + ED
= 9.43 + 14.15
= 23.58 units
Hence, the correct answer is 23.58 units.
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