Question

Two circles touch each other externally at C and AB is common tangent to the circles then $\angle ACB$ is :

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (D)

${90}^{\circ }$

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