Question

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 circle is

• A. $x=4$
• B. $y=2$
• C. $x+\sqrt{3}y=4$
• D. $x+2\sqrt{2}y=6$

Answer 1

The correct option is D $x+2\sqrt{2}y=6$

Common tangent to both the circle is

$B$ divides $C_1:C_2$ in $2:1$ externally
$\therefore B(6,0)$
Let the equation of common tangent be
$\Rightarrow y-0=m(x-6)$
$\Rightarrow mx-y-6m=0$
Tangent to circle of radius $2$
$\therefore \frac{|6m|}{\sqrt{1+m^2}}=2$
$\Rightarrow m=\pm \frac{1}{2\sqrt{2}}$
Equation of tangent $x+2\sqrt{2}y=6$ or $x-2\sqrt{2}y=6$