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Circle and tangient question, AS core 2

  1. May 21, 2006 #1
    Doing a core 2 maths question, realised i cant remember how to find where a tangient and circle meet.

    The circle equation provided in question was x^2 + y^2 -10x + 9 = 0

    same as (x - 5)^2 + y^2 = 4^2

    Question was:

    "Given that line l with gradient 7/2 is a tangient to the circle, and that l touches circle at point T

    find an equation that passes through the centre of the circle and T"

    i tried to find out where line and circle met but wasnt able too. In mark scheme they had a very easy way to do it (m1m2=-1 so gradient is -2/7 and you know the co-ordinates of the centre of the circle, so you use y-y1 = m(x-x1) )

    So, anyway i tried to put line and circle together and realised i couldnt... this is what i did how do you do it?

    What i first was say that the forumula of the straight line is 7/2x + c = y where c is a constant

    i then substituted that in the circle forumula to get (x-5)^2 + (7/2x + c)^2 = 4^2

    multiplied out to get (53/4)x^2 + (7c - 10)x + 9 + c^2 = 0

    since there can only be one result, b^2 - 4ac = 0 so

    (7c -10)^2 - 4 * (53/4) * (9 + c^2) = 0

    which ends up with

    102c^2 -140c + 577 = 0

    which does not have a result... :confused:
  2. jcsd
  3. May 21, 2006 #2


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    HINT: The radius of a circle is always perpendicular to the tangent. What can you say about the gradient two perpendicular lines?

  4. May 21, 2006 #3
    Thats the method they used in the mark scheme:

    so gradient is -2/7, midpoint of circle is 5,0 and you get y = -2/7(x-5)

    But i realised that i didnt know how to get the equation of the original line or where it meets the circle, so i thought i should ask about that here (i really shouldnt have confused it by keeping original question in my post)
    Last edited: May 21, 2006
  5. May 21, 2006 #4


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    Sorry, a missunderstanding on my part. Okay, you now have the equation of the line by using, y - y1 = m(x - x1). Now, if a line and a curve intersect their x and y co-ordinates must be equal at that point. Can you go from here?

  6. May 21, 2006 #5
    ahhh i get it now you do the line going through the centre of the circle... wont bother posting calculation but:

    x = 5 +/- (784/53)^0.5

    thanks for the help!
    Last edited: May 21, 2006
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