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Trignometry

 Joined: 19 Oct 2006 Post: 1558
18 Nov 2006 03:16:23 IST
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Contest [swordfish #4]: Find equation of tangents' mid point
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

From a point P (x1 , y1 ), tangents PA and PB are drawn to the circle x2 +y2 +2gx +2fy +c = 0
If P' and Q' are mid points of PA and PB respectively, find the equation of P'Q'.

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Joined: 18 Dec 2006
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31 Dec 2006 11:45:14 IST
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this problem consists of four small steps:
1)finding the equation of AB
2)finding M(the intersection point of AB and  OP)
3)finding the mid point of PM
4)finding the equation of P'Q'
every step is a short step.
in the first step, let A be (x,y) and C be centre of OP. AC=CP completes first step
infourth step, slope and a point are known.

answer comes out to be x(x1+g)+y(y1+f)=PA/2

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31 Dec 2006 19:34:21 IST
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The Ans is 2x-x1+g=0

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31 Dec 2006 19:40:07 IST
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If c9-g,-f) is centre of circle,
First take c(-g,-f)
Then the middle point of CP, which is P' [(x1-g)/2,(y1-f)/2]
now the equation of PQ is the x = (x of the point P')
So, the equation is  x=(x1-g)/2.
So, 2x- x1+g=0

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6 Jan 2007 19:53:31 IST
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D EQN OF P'Q' IS 2x2 -x1 +g=0

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20 Jan 2007 19:55:46 IST
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swordfish #4 result I'm afraid, no student has been able to answer this question with the correct technique.

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24 Jan 2007 14:38:51 IST
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LINE JOINING P'Q' IS PARALLEL TO CHORD OF CONTACT
EQN OF CHORD OF CONTACT IS xx'+yy'+g(x+x')+f(y+y')+c=0
=)                 x(x'+g)+y(y'+f)+gx'+fy'+c=0
:.P'Q'=     x(x'+g)+y(y'+f)+k=0

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10 Mar 2007 21:55:38 IST
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LINE JOINING P'Q' IS PARALLEL TO CHORD OF CONTACT
equation of chord of contact is :-
xx'+yy'+g(x+x')+f(y+y')+c=0
x(x'+g)+y(y'+f)+gx'+fy'+c=0

so the slope of the P'Q' will be = -(x'+g)/(y'+f)
centre C=(-g,-f)
as  the mid -point ((-g+x')/2,(-f+y')/2)of the line joining CP lies on the line P'Q'
thus the eqn of P'Q' will be by point-slope form
2x(x'+g)+2y(y'+f)+g^2+f^2-x'^2-y'^2=0

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10 Mar 2007 23:54:25 IST
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ankur is correct!!

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11 Mar 2007 01:21:38 IST
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ankur is absolutely correct..

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5 Jun 2007 16:41:06 IST
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from the question for pt. P (X1 Y1)

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5 Jun 2007 16:49:57 IST
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LINE JOINING P'Q' IS PARALLEL TO CHORD OF CONTACT
equation of chord of contact is :-
xx'+yy'+g(x+x')+f(y+y')+c=0
x(x'+g)+y(y'+f)+gx'+fy'+c=0
now we can find the   distance of chord of contact to point P. from similiarity of triangle we will find that the   distance of P'Q' from P will be half of  distance of chord of contact to point P. so distance from P'Q' to P can be known. now we know slope of P'Q' & distance between both parrallel lines
so slope is known & parallel distance is known so we can find the eqn. of P'Q' .

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27 Aug 2007 11:57:00 IST
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x =2 u=78  plzz helpm-e m=90 p=78

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25 Dec 2007 13:38:36 IST
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it is same as the equation of chord AB......

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30 Apr 2008 20:38:49 IST
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THE ans is Ti - S1/2

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27 Nov 2008 08:57:00 IST
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it can be find outr by solving for54

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6 Aug 2010 20:08:23 IST
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18 May 2012 11:21:03 IST
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Contest [swordfish #4]: Find equation of tangents' mid point

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