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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(sqrt3,1)`. A straight line `L`, perpendicular to `PT` is a tangent to the circle `(x-3)^2+y^2=1` then find a common tangent of the two circles

A

`x=4`

B

`y=2`

C

`x+sqrt(3) y =4`

D

`x+2 sqrt(2) y =6`

Text Solution

Verified by Experts

The correct Answer is:
D
The point of intersection of direct common tangnets is `(6,0)`

So, let the equation of common tangnet be
`y-0 = m(x-6)`
As it touches `x^(2)+y^(2) =4`, we have
`|(0-0-6m)/(sqrt(1+m^(2)))|=2`
or `9m^(2)=1+m^(2)`
or `m= +-(1)/(2sqrt(2))`
So, the equation of common tangents are
`y=(1)/(2sqrt(2)) (x-6) ,y= - (1)/(2sqrt(2)) (x-6)` , also `x=2`
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