# Thread: includes circle,chord and tangent concepts

1. ## includes circle,chord and tangent concepts

prove that chord of contact of the pair of tangents to the circle $x^2 + y^2 = 1$ drawn from any point on the line $2x + y = 4$ passes through a fixed point.also,find the co-ordinates of that point.

i could not start the problem

2. Originally Posted by grgrsanjay
prove that chord of contact of the pair of tangents to the circle $x^2 + y^2 = 1$ drawn from any point on the line $2x + y = 4$ passes through a fixed point.also,find the co-ordinates of that point.

i could not start the problem
1. I've made a sketch of the situation (see attachment).
D and E are the tangent points of the tangents from A to the circle.

2. For a start have a look here: Pole and polar - Wikipedia, the free encyclopedia

3. i need to prove this theoritically

4. Did you even look at the websited linked to by earboth? Lots of "theory" there!

5. i could not understand still how to solve problem......
i could not the theory even sorry

6. Originally Posted by grgrsanjay
i could not understand still how to solve problem......
i could not the theory even sorry

Maybe you will be allowed to quote certain theorems.
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.

7. Once you have a suitable proof that all points on the line
form pairs of tangents to the circle, whose chords have a common point of intersection
within the circle, you can find that point using similar triangles.

8. Alternatively,

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 $x_qx+y_qy=1$ for $\left(x_q,\ y_q\right)=(1.6,\ 0.8)=\left(\frac{8}{5},\ \frac{4}{5}\right)$

$\frac{8}{5}x+\frac{4}{5}y=1\Rightarrow\ 8x+4y=5$

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 $\left(x_1,\ y_1\right)$ 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.

$x_1x+y_1y=1,\;\;\;\;for\;\;y_1=-2x_1+4$

$x_1x+\left(-2x_1+4\right)y=1$

$\displaystyle\ y=\frac{1-x\left(x_1\right)}{4-2x_1}$

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.

$y=\frac{1}{2}x$

$\displaystyle\frac{1}{2}x=\frac{1-x\left(x_1\right)}{4-2x_1}$

$\displaystyle\frac{1}{2}x\left(4-2x_1\right)=1-x_1x\Rightarrow\ 2x-x_1x=1-x_1x$

$2x-1=0\Rightarrow\ x=\frac{1}{2}$

Therefore, all the chords have a common point of intersection.