Question

The circles having radii `r_1a n dr_2` intersect orthogonally. The length of their common chord is `(2r_1r_2)/(sqrt(r1 2+r1 2))` (b) `(sqrt(r1 2+r1 2))/(2r_1r_2)` `(r_1r_2)/(sqrt(r1 2+r1 2))` (d) `(sqrt(r1 2+r1 2))/(r_1r_2)`
A. `(2r_(1)r_(2))/(sqrt(r_(1)^(2)+r_(2)^(2)))`
B. `(sqrt(r_(2)^(2)+r_(1)^(2)))/(2r_(1)r_(2))`
C. `(r_(1)r_(2))/(sqrt(r_(1)^(2)+r_(2)^(2)))`
D. `(sqrt(r_(2)^(2)+r_(1)^(2)))/(r_(1)r_(2))`

Answer 1

Correct Answer - 1
Let `/_AB_(1)B_(2)= theta ` Then,
`AD=r_(1)sin theta `( In `Delta ADB_(1))`
and `AD=r_(2)cos theta (ln Delta ADB_(2))`
`:. AD^(2)((1)/(r_(1)^(2))+(1)/(r_(2)^(2)))=1`
or `AD=(r_(1)r_(2))/(sqrt(r_(1)^(2)+r_(2)^(2)))`
Thus, length of common chord `=(2r_(1)r_(2))/(sqrt(r_(1)^(2)+r_(1)^(2)))`