Two circles touch each other externally at P. AB is a direct common tangent touching them respectively at A and B, then ∠APB is:

Given:
Two circles $$C_1$$​ and $$C_2$$​ touch each other externally at P.
AB is their direct common tangent touching $$C_1$$​ at A and $$C_2$$​ at B.
Let the common tangent through P meet AB at T.
Solution:
For circle $$C_1$$​ tangents from T are TA and TP.
$$\Rightarrow$$​ TA = TP $$\quad ...(1)$$​​
For circle $$C_2$$​: tangents from T are TB and TP.
$$\Rightarrow$$​ TB = TP $$\quad ...(2)$$​​
From (1) and (2):
​$$\Rightarrow TA = TB = TP$$​​
Hence, in triangle APB, T is the midpoint of AB and TP = TA = TB.
Now, consider triangle ATP:
Since TA = TP, it is isosceles.
​$$\Rightarrow ∠TPA = ∠TAP = α$$​​
Similarly, in triangle BTP:
TB = TP, so it is isosceles.
​$$\Rightarrow ∠TPB = ∠TBP = β$$​​
Now in triangle APB,
∠APB = ∠TPA + ∠TPB = α + β
Sum of angles in triangle APB:
∠APB + ∠PAB + ∠PBA = 180°
Substituting,
(α + β) + α + β = 180°
2α + 2β = 180°
2(α + β) = 180°
α + β = 90°
Therefore, ∠APB = 90°
Correct answer: (b) 90°