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 circles 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 $x+2\sqrt{2}y=6$

Here, equation of common tangent be

$y=mx\pm 2\sqrt{1+m^2}$
Which is also the tangent to
$(x-3{)}^2+y^2=1$
$\Rightarrow \frac{|3m-0+2\sqrt{1+m^2}|}{\sqrt{m^2+1}}=1\Rightarrow 3m+2\sqrt{1+m^2}=\pm \sqrt{1+m^2}\Rightarrow 3m=-3\sqrt{1+m^2}or3m=-\sqrt{1+m^2}\Rightarrow m^2=1+m^{2}or9m^2=1+m^2\Rightarrow mϵ\varphi orm=\pm \frac{1}{2\sqrt{2}}\therefore y=\pm \frac{1}{2\sqrt{2}}x\pm 2\sqrt{1+\frac{1}{8}}\Rightarrow y=\pm \frac{x}{2\sqrt{2}}\pm \frac{6}{2\sqrt{2}}\Rightarrow 2\sqrt{2}y=\pm (x+6)\therefore x+2\sqrt{2}y=6$