Answer :

Given X and Y are two circles that touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

We have to find ∠APB.

Let ∠CAP = α and ∠CBP = β

CA = CP [The lengths of the tangents from an external point C]

In ΔPAC, ∠CAP = ∠APC = α

Similarly, CB = CP and ∠CPB = ∠PBC = β

We know that sum of interior angles in a triangle is 180°.

Now in ΔAPB,

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ α + β + (α + β) = 180°

⇒ 2α + 2β = 180°

⇒ α + β = 90°

∴ ∠APB = α + β = 90°

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