If line $y=kx$ divides two chords drawn from point $(\sqrt{3},1)$ on the circle $x^2+y^2=4$ in $1:2$ .
Then range of values of $k$
$\bf{My\; Try::}$ Equation of chord $(1)$ which passess through $(\sqrt{3},1)$ and having slope $l$
is $y-\sqrt{3}=l(x-1)$
Now Line $y=kx$ in intersect that chord at $A$. So coordinate of $A$
Similarly Equation of chord $(2)$ which passess through $(\sqrt{3},1)$ and having slope $l$
is $y-\sqrt{3}=m(x-1)$ and line $y=kx$ intersect that line at $B$
Now Solving lines and getting values of $A$ and $B$ and then using Section formula,
But this makes the solution very complicated, My question is how can i solve in easiest way ,
Help required, Thanks
This is really a Euclidean geometry question.
Let $A$ be the point $(\sqrt3,1)$. Let $C$ be the point $(-\sqrt3,1)$ and $B$ the point $(0,-2)$. Take $Y$ as the point which divides the chord $AC$ in the ratio 1:2. Let the line $OY$ meet the (minor) arc $AC$ at $Y'$.
The line $y=kx$ through $Y$ is the line $OY$ through the origin $O$. If this divides another chord $AC'$ in the same way, then $CC'$ must be parallel to $OY$. That is not possible because the tangent at $C$ is parallel to $OY$.
So $C$ is a limiting case. If we take a point $D$ on the (minor) arc $AC$, then it is easy to see that the line $y=kx$ will intersect the arc $AC$ between $A$ and $Y'$ and that there is a second chord $AD'$. Conversely by a suitable choice of $D$ we can get any point on the arc $AY'$.
Similarly, $B$ is the limiting case the other way. So the line $y=kx$ can intersect the circle at any point on the arc $X'Y'$ except for its endpoints. In terms of the gradient, that means that $k$ can be any point of the open interval $(0,\sqrt3)$.