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`

Options

A. `x-sqrt(3)y=1`
B. `x+sqrt(3)y=1`
C. `x-sqrt(3)y=-1`
D. `x+sqrt( 3)y=5`

Answer

A

Explanation

The equation of tangent at P`(sqrt(3), 1)` to the circle `x^(2)+y^(2)=4` is `sqrt(3)x+y=4`. Its slope is `-sqrt(3)`.
So, slope of a line perpendicular to PT is `(1)/(sqrt(3))`.
The equation of tangents of slope `(1)/(sqrt(3))` to the circle `(x-3)^(2)+(y-0)^(2)=1^(2)` are
`y-0= (1)/(sqrt(3))(x-3)pmsqrt(1+((1)/(sqrt(3)))^(2))` or, `sqrt(3)y=x-3 pm 2`
or, `x-sqrt(3)y-1=0 and x-sqrt(3)y-5=0`