What is the distance between the points of tangency of a common tangent AB to two circles I and II which are externally tangent to each other. The radii of the two circles are 9 and 16 units.

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17√3

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Solution

The correct option is A
24

From the diagram above, AC $^{1}$ =16
and BC $^{2}$ =9.
Also C $^{1}$ - C $^{2}$ =25.
PC $^{1}$ = 7 (16-9)
AB=PC $^{2}$
C $^{1}$ - P - C $^{2}$ is a right angled triangle
( $C^{1} - C^{2})^{2}$ = PC $^{12}$ +PC $^{22}$
Thus, PC $^{2}$ = $\sqrt{625 - 49}$
= $\sqrt{576}$ = 24

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What is the distance between the points of tangency of a common tangent AB to two circles I and II which are externally tangent to each other. The radii of the two circles are 9 and 16 units.

The correct option is A 24 From the diagram above, AC1=16 and BC2=9. Also C1 - C2=25. PC1 = 7 (16-9) AB=PC2 C1 - P - C2 is a right angled triangle (C1−C2)2 = PC12+PC22 Thus, PC2 =√625−49 = √576 = 24