A possible equation of is, A tangent is drawn to the circle at the point . A straight line , perpendicular to is a tangent to the circle . [2012]

Question

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A possible equation of $L$ is

Smart Achievers · Accepted Answer

(1)

$Equation of tangent P T to the circle x^{2} + y^{2} = 4$ $at the point P (\sqrt{3}, 1) is$ ${(x - 3)}^{2} + y^{2} = 1$

$Let the line L, perpendicular to tangent P T be$

$x - \sqrt{3} y + λ = 0$

$As it is tangent to the circle {(x - 3)}^{2} + y^{2} = 1$

$∴$ Length of perpendicular from centre of circle to the tangent = radius of circle

$\Rightarrow$ $| \frac{3 + λ}{2} |$ $= 1 \Rightarrow λ = - 1 or - 5$

$∴$ $Equation of L can be$ $x - \sqrt{3} y = 1 or x - \sqrt{3} y = 5$

Solution

Q.
A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A possible equation of $L$ is

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

2 $x + \sqrt{3} y = 1$

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

4 $x + \sqrt{3} y = 5$

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Solution

Q. 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. [2012] Q. A possible equation of L is

1 x-3y=1

2 x+3y=1

3 x-3y=-1