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May 20th, 2013, 08:51 AM  #1 
Senior Member Joined: Sep 2011 Posts: 395 Thanks: 0  Circles and Chords
For this question, I've defined the line equation perpendicular to the circle's radius, and found measurements of the chords of either of the 2 lines, but I'm unsure what to do next. I most likely need a midpoint of the bisector for both of the lines to find their equation but, from what I've got so far, I'm not sure how to do this. Can anyone help, please? Many thanks. Q. is the equation of a circle. 2 lines, L & M, intersect at (2, 2). The distance from the centre of the circle to each line is . Find the equation of L & M. Attempt: Let b = (2, 2), From circle equation: centre c = (1, 2) & radius: If distance from c to b = 5 & distance from c to either line is , then: . 1/2 length of bisector of either line. Thus, chord length = Slope of [bc]: . Perpendicular slope = . Equation of line perpendicular to [bc]: Ans. (From text book): & 
May 20th, 2013, 10:19 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007 
The line through (2, 2) with slope m has equation y + 2  m(x  2) = 0. The distance from the point (1, 2) to this line is 3m + 4/sqrt(m² + 1) = 2?5. Squaring and rearranging leads to (11m  2)(m  2) = 0, so m = 2/11 or 2. Hence the required lines are y  2x + 6 = 0 and y  (2/11)x + 26/11 = 0. (There are various equivalent forms of those equations.) 
May 20th, 2013, 11:06 AM  #3 
Senior Member Joined: Sep 2011 Posts: 395 Thanks: 0  Re: Circles and Chords
Thanks for getting back to me on that. Just one thing; I'm not entirely clear on where you got , and why multiplying it against . Could you explain that part? Thanks again. 
May 20th, 2013, 10:16 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007 
I used this formula.


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