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A tangent PT is drawn to the circle x^(2...

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 possible equation of L is `:`

A

`x-sqrt(3) y =1`

B

`x+ sqrt(3) y =1`

C

`x-sqrt(3) y = -1`

D

`x+sqrt(3) y =5`

Text Solution

Verified by Experts

The correct Answer is:
1
The equation of tangent at `P(sqrt(3),1)` is `sqrt(3)x+y=4`
The slope of line perpendicular to the above tangent is `1//sqrt(3)`. So, the equations of tangents with slope `1//sqrt(3)` to `(x-3)^(2)+y^(2)=1` will be
`y= (1)/(sqrt(3))(x-3)=- 1sqrt(1+(1)/(3))`
or `sqrt(3) y = x-3+- (2)`
i.e., `sqrt(3) y = x-1 ` or `sqrt(3) y = x-5`
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