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Q.

In the given figure, PT is a common tangent to the circles touching externally at P and AB is another common tangent touching the circles at A and B. Find the APB.  

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a

90 o  

b

80 o  

c

70 o  

d

60 o   

answer is A.

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Detailed Solution

Given that PT is a common tangent to the circles touching externally at P and AB is another common tangent touching the circles at A and B.
We have to find APB.  
The tangent drawn at any point of a circle is perpendicular to the radius through the point of contact.
The required figure geometry is shown below,
Question ImageSuppose TAP=α   and TBP=β   . Then, TA = TP, because the lengths of the tangents from an external point is equal. Similarly, TB = TP. Therefore, CAP=APT=α   and TPB=PBT=β.   In the triangle APB by using the angle sum property of the triangle, PAB+PBA+APB = 180 ° α+β+(α+β) = 180 ° 2(α+β) = 180 ° α+β = 90 °   The value is APB= 90 °  
Therefore, the correct option is 1.
 
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