The length of the transversal common tangent to the circle `x^(2)+y^(2)=1` and `(x-t)^(2)+y^(2)=1` is `sqrt(21)`, then t=

Options

A. `pm2`
B. `pm5`
C. `pm3`
D. none of these

Answer

B

Explanation

If d is the distance between the centres of two circles of radii `r_(1)` and `r_(2)` , then
Length of transversal common tangent `=sqrt(d^(2)-(r_(1)+r_(2))^(2))`

Centres of the given circles are `C_(1)(0, 0)` and `C_(2)(t, 0)` and corresponding radii are `r_(1)=1` and `r_(2)=1` respectively.
`:. d=sqrt((t-0)^(2)+0)=|t|`
It is given that :
Length of transversal common tangent `=sqrt(21)`
`rArr sqrt(d^(2)-(r_(1)+r_(2))^(2))=sqrt(21)rArr t^(2)-2^(2)=21 rArr t= pm 5`
`ul("ALITER")` Since given circles are of equal radius. Therefore, R is the mid-point of `C_(1)C_(2)`. Also , `DeltaC_(1) LR ~= DeltaC_(2)MR`.
`:. ` R is the mid-point of LM.
In `DeltaC_(1) LR`, we have
`C_(1)R^(2)=C_(1)L^(2)+LR^(2)`
`rArr LR=sqrt(C_(1)R^(2)-C_(1)L^(2))=sqrt((t^(2))/(4)-1)=sqrt((t^(2)-4)/(2))`
`:. LM = sqrt(21)`
`rArr 2 LR = sqrt(21) rArr sqrt(t^(2)-4)=sqrt(21)rArr t^(2)=25 rArr t = pm 5`