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January 1st, 2013, 05:01 AM  #1 
Newbie Joined: Jun 2011 Posts: 17 Thanks: 0  [CIRCLE] Equation of tgt. and distance
Q 1: Find the equation to the tangent at A to the circle , where the radius through A makes an angle with the xaxis. My attempt: Since it is given that the line through A and centre makes an angle with xaxis, so the slope of normal at A is and hence the slope of tangent at A is .Let .Expanding the equation of circle, we get and tangent to it is . So, or Also from two point form of a straight line, equation of tangent is or Equations (1) and (2) are same. So on comparing coefficients and constants, we obtain Obtaining p and q from these equations looks a tough job. I don't know after that Q 2: A line is drawn through a fixed point to cut the circle at A and B. Then PA X PB is equal to. Attempt: I first thought of the line to pass through the centre(it is one of the cases) and then calculating the point of intersection and then applying the distance formula but , it wasn't a good idea. I only know that PA X PB = PC X PD(refer figure) [attachment=0:3dc7m5za]Untitled.jpg[/attachment:3dc7m5za] 
January 1st, 2013, 02:47 PM  #2 
Senior Member Joined: Jun 2011 Posts: 298 Thanks: 0  Re: [CIRCLE] Equation of tgt. and distance
The center of the circle is at AO and the tangent form, so the slope of the tangent line is The coordinate at A is Now you can find tangent line intersection on the axis. Suppose the result is Suppose the tangent line is . You can find by substituting for and in the equation. 
January 5th, 2013, 06:41 AM  #3  
Newbie Joined: Jun 2011 Posts: 17 Thanks: 0  Re: [CIRCLE] Equation of tgt. and distance Quote:
or First question is now figured out. May i get some help on the second question  
January 10th, 2013, 02:09 AM  #4  
Newbie Joined: Jun 2011 Posts: 17 Thanks: 0  Re: [CIRCLE] Equation of tgt. and distance
Just to tell, its seems that i have found the solution.http://en.wikipedia.org/wiki/Circle#Theorems. Point #2>Tangentsecant theorem. Quote:
So in my case, :P  

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