Question

Two concentric circles having radii 73 and 55 are given. The chord of the circle with larger radius touches the circle with smaller radius. Find the length of the chord.

Answer 1

R , r are the radii of Concentric circles .

It is given that ,

R = 73 ,

r = 55 ,

AB is a tangent to the smaller circle at P.

AB is chord .

OP is perpendicular to AB .

In OPB ,

<OPB = 90° ,

OB² = OP² + PB²

R² = r² + PB²

PB² = R² - r²

= 73² - 55²

= ( 73 + 55 )( 73 - 55 )

= 128 × 18

= 4 × 4 × 4 × 4 × 3 × 3

PB = 4 × 4 × 3

PB = 48

Chord = AB

= 2 × PB

= 2 × 48

= 96

I hope this helps you.

: )

Answer 2

Let a and b is the radius of circle.

First see figure which is attached.

Now, apply Pythagoras theorem in ΔAOP

$OP^{2} +AP^{2} =OA^{2}$

$55^{2} +AP^{2} =73^{2}$

$AP^{2} =73^{2} -55^{2}$

$AP^{2} =2304$

$AP=\sqrt{2304}$

$AP = 48$

AP= PB

Hence, length of chord is 2AP = 96.