Question

Two cirele drawn with centres A and B touch each other externally at C, O is a point on the tangent drawn at C, OD and OE are tangents drawn to the two circles of centres A and I respectively. If ∠COD = 56°, ∠COE = 40°, ∠ACD = x° and ∠BCE = y°. Let us Prove that OD = OC = OE and y - x = 8.​

Answer 1

Answer:

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

To find : ∠ACB

Proof:

Let P be a point on AB such that, PC is at right angles to the Line Joining the centers of the circles.

Note that, PC is a common tangent to both circles.

This is because tangent is perpendicular to radius at point of contact for any circle.

let ∠PAC= α and ∠PBC = β.

PA = PC [lengths of the tangents from an external point C]

In a triangle CAP, ∠PAC = ∠ACP = α

similarly PB = CP and ∠PCB = ∠CBP = β

now in the triangle ACB,

∠CAB + ∠CBA + ∠ACB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180° (Since ∠ACB = ∠ACP + ∠PCB = α + β.

2α + 2β = 180°

α + β = 90°

∴ ∠ACB = α + β = 90°

the answer will be option D

Step-by-step explanation:

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