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Finding Equation of Circle

  1. Sep 16, 2008 #1
    1. Find the equation of a circle tangent to both axis with its center in the 4th quadrant passing through (8,-9)



    2.Equation of Circle (X-a)^2+(Y-b)^2)=r^2



    3. I tried solving graphically, but am not sure whether my graph is precise

    Thanks for the help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2008 #2

    HallsofIvy

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    Think about the fact that the circle is "tangent to both axis ". If a circle has center at (a, b) and is tangent to the y-axis, then the line from (a, b) to (0, b) (the point of tangency) is a radius- and has length a. If a circle has center at (a, b) and is tangent to the y-axis, then its radius is a.

    If a circle has center at (a,b) and is tangent to the x-axis, then the line from (a, b) to (a, 0) (the point of tangency) is a radius- and has length b. If a circle has center at (a, b) and is tangent to the x-axis, then its radius is b.

    Now, what if a circle has center at (a, b) and is tangent to both axes? Then you get that the radius is r= a= b. That is, you know that the general equation of a circle, (x-a)2+ (y- b)2= r2, for this case becomes (x-a)2+ (y- a)2= r2 so you only need to find a.

    You are told that the circle passes through (8, -9) so you know that x= 8, y= -9 satisfies that equation. Put x= 8, y= -9 and solve for a. You will get a quadratic equation for a so there may be 2 correct answers.
     
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