A tangent P T is drawn to circle x^2+y^2=4 at point P(√{3}, 1) . A straight line L , perpendicular

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10-1.Circle and System of Circles

normal

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 $(x-3)^2+y^2=1$.

$1.$ A common tangent of the two circles is

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

$2.$ A possible equation of $L$ is

$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

Give the answer question $1$ and $2.$

$(D,A)$

$(B,D)$

$(B,C)$

$(C,D)$

(IIT-2012)

Solution

$1.$ $Image$

Equation of tangent at $(\sqrt{3}, 1)$

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

$Image$

B divides $C _1 C _2$ in $2 : 1$ externally

$\therefore B(6,0)$

Hence let equation of common tangent is

$y-0=m(x-6) $

$m x-y-6 m=0$

length of $\perp^{\text {r }}$ dropped from center $(0,0)=$ radius

$\left|\frac{6 m}{\sqrt{1+m^2}}\right|=2 \Rightarrow m= \pm \frac{1}{2 \sqrt{2}}$

$\therefore$ equation is $x+2 \sqrt{2} y=6$ or $x-2 \sqrt{2} y=6$

$2.$ Equation of $L$ is

$x-y \sqrt{3}+c=0$

length of perpendicular dropped from centre $=$ radius of circle

$\therefore\left|\frac{3+C}{2}\right|=1 \Rightarrow C=-1,-5 $

$\therefore x-\sqrt{3} y=1 \text { or } x-\sqrt{3} y=5$

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