Question : AB is the common tangent to both circles as shown in the given figure. What is the distance between the centres of the circles?
Option 1: 20 cm
Option 2: 15 cm
Option 3: 10 cm
Option 4: 30 cm
Correct Answer: 30 cm
Solution :
Given that AB is the common tangent to both circles.
In the circle with centre C, $\angle$CAE = 90°
In the circle with centre D, $\angle$DBA = 90°
and $\angle$AEC = $\angle$BED (vertically opposite angle)
So, $\triangle$CAE $\sim$ $\triangle$DBC
⇒ $\frac{CA}{AE}=\frac{DB}{BE}$
⇒ $\frac{4}{3}=\frac{DB}{15}$
⇒ DB = $\frac{4×15}{3}$ = 20 cm
In $\triangle$CAE,
⇒ CE
2
= AC
2
+ AE
2
⇒ CE
2
= 4
2
+ 3
2
= 5
2
⇒ CE = 5 cm
In $\triangle$DBE,
⇒ DE
2
= DB
2
+ BE
2
⇒ DE
2
= 20
2
+ 15
2
= 25
2
⇒ DE = 25 cm
$\therefore$ The distance between the centres of the circle = CE + DE = 5 + 25 = 30 cm
Hence, the correct answer is 30 cm.
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