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Equation of a Circle

  1. May 15, 2013 #1
    I'm having some difficulty with this question. Can anyone help me out, please?

    Many thanks.

    1. The problem statement, all variables and given/known data

    A circle of radius length [itex]\sqrt{20}[/itex] contains the point (-1, 3). Its centre lies on the line x + y = 0. Find the equations of the 2 circles that satisfy these conditions.

    2. Relevant equations

    3. The attempt at a solution

    Circle equation for the point: [itex](-1-h)^2+(3-k)^2=20\rightarrow h^2+k^2+2h-6k-10=0[/itex]
    Because x+y=0 contains centre c as (-g, -f), we can say -g - f = 0.
    Thus, -g = -2h & -f = 6k.
    And so, [itex]x+y=0\rightarrow -2h+6k=0\rightarrow h=3k[/itex].
    Returning to circle equation: [itex](3k)^2+k^2+2(3k)-6k-10=0\rightarrow 10k^2=10\rightarrow k=+/- 1[/itex]
    For k = 1: [itex]h=3k\rightarrow h=3[/itex].
    Thus, centre c = (3, 1) and equation = [itex](x-3)^2+(y-1)^2=20\rightarrow x^2+y^2-6x-2y-10=0[/itex]
    For k = -1: [itex]h=-3[/itex]
    Thus, centre c = (-3, -1) and equation = [itex](x+3)^2+(y+1)^2=20\rightarrow x^2+y^2+6x+2y-10=0[/itex]

    Ans.: (From text book): [itex]x^2+y^2-2x+2y-18=0[/itex] & [itex]x^2+y^2+10x-10y+30=0[/itex]
     
  2. jcsd
  3. May 15, 2013 #2

    CAF123

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    Gold Member

    You can't have a centre of (3,1) for example because the centre must lie on the line x=-y, so in your notation the centre is like (h,-h). The condition you want is that the distance between the centre and the point is less or equal to ##\sqrt{20}##. That is, $$\sqrt{(h+1)^2 + (k-3)^2} \leq \sqrt{20}$$ Use this and the condtion that h=-k, h and k denoting the coordinates of the centre.

    EDIT: I misinterpreted what 'contained' meant - I think it should be that the point lies on the circumference of the circle so there is a strict equality in the above expression.
     
    Last edited: May 15, 2013
  4. May 15, 2013 #3
    Ok, I have it now. Thank you very much.
     
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