Question

a chord AB of a circle C1 of radius root 3 + 1 cm touches a circle C2 of radius root 3 - 1 cm then the length of AB is

Answer 1

Answer:

The length of AB is $4\sqrt[4]{3}$ cm.

Step-by-step explanation:

We are given two circle C1 and C2. Please see attachment for diagram.

Radius of circle $C_1=\sqrt3+1$ cm

Radius of circle $=C_2=\sqrt3-1$ cm

A chord of circle C1 touches a circle C2.

AB must be tangent of circle C2.

In triangle OMB, OM⊥AB because radius perpendicular to tangent of circle.

where, OM=Radius of circle C2 and OB is radius of circle C1

$OM=\sqrt3-1$ cm

$OB=\sqrt3+1$ cm

In ΔOMB, ∠OMB=90°

$MB^2=OB^2-OM^2$                   By Pytahgoreous theorem

$MB^2=(\sqrt3+1)^2-(\sqrt3-1)^2$

$MB^2=3+1+2sqrt3-3-1+2sqrt3$

$MB^2=4\sqrt3$

$MB=2\sqrt[4]{3}$

AB=2MB = $4\sqrt[4]{3}$

Hence, The length of AB is $4\sqrt[4]{3}$ cm.