Question

A tangent PT is drawn to the circle `x^2 + y^2= 4` at the point `P(sqrt3,1)`. A straight line L is perpendicular to PT is a tangent to the circle `(x-3)^2 + y^2 = 1` Common tangent of two circle is: (A) `x=4` (B) `y=2` (C) `x+(sqrt3)y=4` (D) `x+2(sqrt2)y=6`
A. `x-sqrt3y=1`
B. `x+sqrt3y=1`
C. `x-sqrt3y=-1`
D. `x+sqrt3y=5`

Answer 1

Here, tangent to `x^(2)+y^(2)=4 "at" (sqrt3,1) "is" sqrt3x+y=4" "...(i)`
AS, L is perpendicular to `sqrt3x + y = 4`
`rArrx-sqrt3y=lambda` which is tangent to
`(x-3)^(2)+y^(2)=1`
`rArr (|3-0-lambda|)/(sqrt(1+3))=1`
`rArr |3-lambda|=2`
`rArr 3-lambda=2,-2`
`therefore lambda=1,5`
`rArrL:x-sqrt3y=1, x=-sqrt3y=5`