A tangent PT is drawn to the circle x2+y2=4 at the point P(3,1). A straight line L, perpendicular to PT is a tangent to the circle (x−3)2+y2=1. A possible equation of L is

An equation of PT is 3x+y=4. (1) ⇒ equation of straight line perpendicular to PT is 3y−x=c (2) It will be tangent to (x−3)2+y2=1 if and only if length of perpendicular from (3,0) to (2) = 1 ⇔|0−3−c3+1|=1

A tangent PT is drawn to the circle x2+y2=4 at the point P(3,1). A straight line L, perpendicular to PT is a tangent to the circle (x−3)2+y2=1. A possible equation of L is

An equation of PT is 3x+y=4. (1) ⇒ equation of straight line perpendicular to PT is 3y−x=c (2) It will be tangent to (x−3)2+y2=1 if and only if length of perpendicular from (3,0) to (2) = 1 ⇔|0−3−c3+1|=1

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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 ${(x - 3)}^{2} + y^{2} = 1$ . A possible equation of L is

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$x - \sqrt{3} y = 1$

$x + \sqrt{3} y = 1$

$x - \sqrt{3} y = - 1$

$x + \sqrt{3} y = 5$

answer is A.

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Detailed Solution

An equation of PT is $\sqrt{3} x + y = 4$ . (1)
$\Rightarrow$ equation of straight line perpendicular to PT is $\sqrt{3} y - x = c$ (2)
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It will be tangent to ${(x - 3)}^{2} + y^{2} = 1$ if and only if length of perpendicular from $(3, 0) t o (2) = 1$
$\Leftrightarrow | \frac{0 - 3 - c}{\sqrt{3 + 1}} | = 1$