Correct Answer - 2
We have `sin phi = d//r_(1)` and `cos phi = d//r_(2)`. ( where 2d is the length of the comon chord ). Then,
`l=(d^(2))/(r_(1)^(2))+(d^(2))/(r_(2)^(2))`
or `d=(r_(1)r_(2))/(sqrt(r_(1)^(2)+r_(2)^(2)))`
or `2d=(2r_(1)r_(2))/(sqrt(r_(1)^(2)+r_(2)^(2)))= (24)/(5),` where `r_(1)=3`
or `d=(6r_(2))/(sqrt(9+r_(2)^(2)))=(24)/(5)`
or `r_(2)=4`
From the figure,
`sin. (theta)/(2)=(r_(2)-r_(1))/(C_(1)C_(2))`
where `C_(1)^(2)C_(2)^(2)=r_(1)^(2)+r_(2)^(2)`
or `C_(1)C_(2)=5`
`:. sin. (theta)/(2)=(1)/(5)`
or `cos . (theta)/(2)=(sqrt(24))/(5)`
`:. sin theta = 2 x (1)/(5) xx (sqrt(24))/(5)`
`=(4 sqrt(6))/(25)`
or `theta = sin^(-1). (4 sqrt(6))/(25)`
Also, `AB = sqrt(C_(1)C_(2)^(2)-(r_(1)-r_(2))^(2))= sqrt(25-1)=sqrt(24)`