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Tangent to the centrum edge of a circle

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi i have a circle that is shown by (x-7)2+(y+1)2=20

    i also have a line y=2x-5 and i have to explain why the line is a tangent to the edge of the circle

    i know that the circle has the centre in (7,1) and that the radius of it is 4,4

    2. Relevant equations

    i know i have to use the formular G =l ad+be+c l/√a2+b2

    where the result should be something like the centre +4,4

    3. The attempt at a solution

    i have tried to change my y=2x-5 so it stands in the form ax+by+c=0 and i think it will end up like 2x+1y+(-5)=0 but i really dont know what to do next. I have attached a photo of what i mean with the tangent. the one that i am speaking about is the red
     

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  2. jcsd
  3. Jan 6, 2013 #2

    mfb

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    You don't have to. And if you want to use it, you should find out what it does, and what a,b,c,d and e are.

    y=2x-5 is not the same as 2x+1y+(-5)=0, there is a sign error.

    Do you know how to compute the distance between a line and a point?
     
  4. Jan 6, 2013 #3

    haruspex

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    That's the distance from the line ax+by+c=0 to the point (d,e), right?
    What point do you think you should plug in?
    Not quite. You have a sign wrong.
     
  5. Jan 6, 2013 #4
    sorry i am not sure if i understand what you mean with what point i should plug in

    if you mean what i think i should write in the formular then i believe it is G =l -2*7+0*1+5 l/√a-2+02

    but that just comes out wrong for a reason.

    with asking if i know how to compute the answer if it means if i know how to use a computer to solve it i already tried. My problem is that the computer shows me that the centre of the circle is in 7,-1 instead which i dont understand
     
  6. Jan 6, 2013 #5

    TSny

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    I'm not familiar with your G formula. Are you required to use it?

    If not, you could approach the problem by trying to find where the line intersects the circle. There are three possibilities: (1) the line intersects the circle in two distinct points (2) the line intersects the circle at only one point (3) the line doesn't intersect the circle at all.
     
  7. Jan 6, 2013 #6
    but that is also what the formular i had shows.

    it shows you the distance between the centre of the circle and the line. Then you will know if that distance is bigger than the radius it doesn't intersect the circle if its smaller than the radius it means it does intersect it two times and if it is the same then that would be one time and also the right answer

    how do you do it?
     
  8. Jan 6, 2013 #7

    TSny

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    OK. I didn't know what the symbols in your G equation stood for. Ignore my post and carry on!
     
  9. Jan 6, 2013 #8

    TSny

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    To find where the line intersects the circle, you solve the circle and line equations simultaneously. The line equation says y = 2x-5. You could substitute this expression for y into the circle equation and try to solve the resulting equation for x.
     
  10. Jan 6, 2013 #9
    so you mean i just have to isolate x in the circle by putting 2x-5 in the place of y and then find x?

    i just did and it says 5 over 2.6 ...thats not right it should be 4,4
     
  11. Jan 6, 2013 #10

    TSny

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    Yes, that's how you do it.
    You wouldn't expect x to be 4.4. But it's not 5/2.6 either. If you post your steps we can look at it.
     
  12. Jan 6, 2013 #11

    SteamKing

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    The center of the circle is not at (7,1) according to the equation you have given in the OP.
     
  13. Jan 6, 2013 #12

    haruspex

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    The formula you quote is for the distance from a line to a point. You are trying to show that this line is a tangent to a circle of known centre and radius. So for what point would it be useful to know its distance from the line?
    Looks like you tried plugging in the correct point (the centre of the circle) but I can't tell whether you did it correctly because I cannot tell what signs you used. Also, as I mentioned, you had a sign wrong in converting the eqn to ax+by+c=0. So to make sure we have things right, please list what values you are filling in for a, b, c, d and e. (The '0' you have in the formula for G is definitely wrong. None of them are 0.)
     
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