Question

Two circles touch each other externally at P .AB is common tangent to the circles touching them at A nd B. the value of <APB is.

Answer 1

Hey there,
Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.To find : ∠APB
Proof: let ∠CAP = α and ∠CBP = β.
CA = CP [lengths of the tangents from an external point C]
In a triangle PAC, ∠CAP = ∠APC = α
similarly CB = CP and ∠CPB = ∠PBC = β
now in the triangle APB,
∠PAB + ∠PBA + ∠APB = 180°   [sum of the interior angles in a triangle]
α + β + (α + β) = 180°
2α + 2β = 180°
α + β = 90°
∴ ∠APB = α + β = 90°
Hope this helps!

Answer 2

Answer:

Hey there,

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

Proof: let ∠CAP = α and ∠CBP = β.

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

In a triangle PAC, ∠CAP = ∠APC = α

similarly CB = CP and ∠CPB = ∠PBC = β

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180°   [sum of the interior angles in a triangle]

α + β + (α + β) = 180°

2α + 2β = 180°

α + β = 90°

∴ ∠APB = α + β = 90°

Hope this helps!

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