Discussion Forum

1)

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 thye circle $(x-3)^{2}+y^{2}=1$

A possible equation of L is 

Answer:

Option A

Explanation:

(i) Equation of a tangent to at (x₁,y₁) is 

x²+y²=r²

 

at (x₁,y₁) is 

xx₁+yy₁=r²

(ii)If ax+by+c =0 is tangent to (x-h)²+(y-k)²=r²

|cp|=r

Here tangent to $x^{2}+y^{2}=4$ at the point

$P(\sqrt{3},1)$  is $\sqrt{3} x+y=4$.............(i)

 As. L is perpendicular to $\sqrt{3} x+y=4$

 $\Rightarrow$  $x-\sqrt{3}y= \lambda$

 which is tangnet to $(x-3)^{2}+y^{2}=1$

 $\Rightarrow$   $\frac{|3-0-\lambda|}{\sqrt{1+3}}=1$

$\Rightarrow$  $|3-\lambda|=2$

$\Rightarrow$  $3-\lambda=2,-2$

  $\therefore$  $\lambda=1,5$

 $\Rightarrow$  $L:x- \sqrt{3} y=1, x-\sqrt{3}y=5$