# includes circle,chord and tangent concepts

• Sep 28th 2010, 08:29 AM
grgrsanjay
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
• Oct 2nd 2010, 11:35 PM
earboth
Quote:

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
• Oct 4th 2010, 06:22 AM
grgrsanjay
i need to prove this theoritically
• Oct 4th 2010, 07:29 AM
HallsofIvy
Did you even look at the websited linked to by earboth? Lots of "theory" there!

earboth's picture was to help you get started, not give you the answer.
• Oct 9th 2010, 07:19 AM
grgrsanjay
i could not understand still how to solve problem......
i could not the theory even sorry

• Oct 12th 2010, 05:35 AM
Quote:

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.
• Oct 12th 2010, 03:17 PM
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.
• Oct 13th 2010, 04:34 PM
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.