A tangent PT is drawn to the circle `x^(2)+y^(2)=4` at the point `P( sqrt(3),1)`. A straight line L, perpendicular to PT , is a tangent to the circle `(x-3)^(2)+y^(2)=1`
A common tangent of the two circles is

Options

A. x = 4
B. y = 2
C. `x + sqrt3y = 4`
D. `x+ 2 sqrt2 y = 6`

Answer

D

Explanation

Here, equation of common tangent be

`y=mxpm2sqrt(1+m^2)`
which is also the tangent to
`(x-3)^(2)+y^(2)=1`
`rArr (|3m-0+2sqrt(1+m^(2))|)/(sqrt(m^(2)+1))=1`
`rArr 3m+2sqrt(1+m^(2))=pmsqrt(1+m^(2))`
`rArr 3m=-3sqrt(1+m^(2))`
`or 3m=-sqrt(1+m^(2))`
`rArr m^(2)=1+m^(2)or 9m^(2)=1+m^(2) `
`rArr m inphiorm=pm(1)/(2sqrt2)`
`therefore y=pm(1)/(2sqrt2)xpm2sqrt(1+(1)/(8))`
`rArr y=pm(x)/(2sqrt2)pm(6)/(2sqrt2)`
`rArr 2sqrt2y=pm(x+6)`
`therefore x+2sqrt2y=6`