Tangents are drawn to the circle $x^2+y^2=a^2$ from a poiny which always lies on the line $lx+my=1$. Prove that the locus of the mid point of the chords of contact is $x^2+y^2-a^2(lx+my)=0$.
My Attempt: Given circle is $x^2+y^2=a^2$ Let it be equation $(1)$.
Again, Given line is $lx+my=1$ Let it be equation $(2)$.
Let $(h,k)$ be a point on $lx+my=1$. So, $lh+mk=1$-----$(3)$.
Please help me to solve from here.
Just a hint: We know that the equation of the chord of contact of tangents from the point $(h,k)$ to our circle is $$hx+ky=a^2 \tag {a } $$ and let $(\alpha, \beta)$ be the mid point of the chord, then the equation of the chord we know is $$x\alpha+y\beta = \alpha^2 +\beta^2 \tag {b} $$ But however equations $(a) $ and $(b)$ represent the same. So solve for $h $ and $k $ from $(a)$ and $(b) $ and substitute in the equation $(3) $ and the result follows. Hope it helps.