<s>[INST] Tangents are drawn to the circle $x^2+y^2=a^2$ from a point which always lies on the line $lx+my=1$. Tangents are drawn to the circle $x^2+y^2=a^2$
from a point which always
lies on the line $lx+my=1$. Prove that the locus of
the mid point of
chords of contact is $x^2 + y^2 -a^2(lx+my)=0$
My Attempt::
let mid point of chord of contact be $(h, k)$ &
point $(x1, y1)$ be always
lying on line: $lx+my=1$ then we have
$lx_1+my_1=1$ ...(1)
Now equation of chord of contact to circle:
$x^2+y^2=a^2$ is $xx_1+yy_1=a^2$ but
$(h, k)$ lies on chord of contact hence satisfying
equation of chord gives us
$hx1+ky1=a^2$ .......(2)
Please help me to complete this proof 
.
 [/INST] 
You can write the equation of the polar line (green line in the figure):
$$x_1x+y_1y=a^2$$
where $(x_1,y_1)$ is a point of the line $lx+my=1$ (red line).
The equation of the normal (blue line) to the polar line, passing through $(x_1,y_1)$ is:
$$y={y_1\over x_1}x$$
so we can find the middle point $M$ of the chord:
\begin{cases}x_1x+y_1y=a^2\\y={y_1\over x_1}x\end{cases}
Solving this system, we get:
$$M\left(x_M={a^2x_1\over x_1^2+y_1^2},y_M={a^2y_1\over x_1^2+y_1^2}\right)$$
Now we must eliminate $x_1$ and $y_1$, thus:
\begin{cases}x_M={a^2x_1\over x_1^2+y_1^2}\\y_M={a^2y_1\over x_1^2+y_1^2}\\lx_1+my_1=1\end{cases}
Squaring and adding the first and the second equation:
\begin{cases}x_M^2+y_M^2=a^4{x_1^2+y_1^2\over (x_1^2+y_1^2)^2}={a^4\over x_1^2+y_1^2}\\x_M={a^2x_1\over x_1^2+y_1^2}\\y_M={a^2y_1\over x_1^2+y_1^2}\\lx_1+my_1=1\end{cases}
\begin{cases}{x_M^2+y_M^2\over a^2}={a^2\over x_1^2+y_1^2}\\x_M={x_M^2+y_M^2\over a^2}x_1\longrightarrow a^2x_M=(x_M^2+y_M^2)x_1\\y_M={x_M^2+y_M^2\over a^2}y_1\longrightarrow y_M={x_M^2+y_M^2\over a^2}{(1-lx_1)\over m}\longrightarrow a^2my_M-(x_M^2+y_M^2)=-(x_M^2+y_M^2)lx_1\\y_1={1-lx_1\over m}\end{cases}
so $a^2my_M-(x_M^2+y_M^2)=-a^2lx_M\longrightarrow x_M^2+y_M^2-a^2(lx_M+my_M)=0$.
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