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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 is perpendicular to PT is a tangent to the circle `(x-3)^2 + y^2 = 1` Common tangent of two circle is:

A

`x-sqrt3y=1`

B

`x+sqrt3y=1`

C

`x-sqrt3y=-1`

D

`x+sqrt3y=5`

Text Solution

Verified by Experts

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`
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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 then find a common tangent of the two circles

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Knowledge Check

  • 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

    A
    `x=4`
    B
    `y=2`
    C
    `x-sqrt(3)y=5`
    D
    `x+2sqrt(2)y=6`

  • 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 solution of L is

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

  • If y = mx + c is a tangent to the circle (x – 3)^2 + y^2 = 1 and also the perpendicular to the tangent to the circle x^2 + y^2 = 1 at (1/sqrt(2),1/sqrt(2)) , then

    A
    `c^2 + 6c + 7 = 0`
    B
    `c^2 - 6c + 7 = 0`
    C
    `c^2 + 6c – 7 = 0`
    D
    `c^2 – 6c – 7 = 0`

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