prove that chord of contact of the pair of tangents to the circle drawn from any point on the line passes through a fixed point.also,find the co-ordinates of that point.
i could not start the problem
Your problem can be more easily solved then.
The .pdf attachment shows one way of seeing the geometry.
I doubt somehow that you need to go to those lengths,
but hopefully it is of some help in spite of being general.
the point Q is closest to the circle.
The line perpendicular to y=-2x+4, which passes through the circle centre is 2y=x.
The point of intersection of 2y=x and y=-2x+4 is (1.6, 0.8).
The equation of the chord of contact from Q is for
The point of intersection of 2y=x and 8x+4y=5 gives us the co-ordinates of the point we are looking for.
Also, if we pick an arbitrary point on the line y=-2x+4,
then we can write a general equation for all of the chords of contact from points on this line to the circle.
Since all of these chords of contact have different slopes,
if they all intersect 2y=x at a common point, then the proof is complete.
Therefore, all the chords have a common point of intersection.