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Equations of circles tangent to graph

  1. Jun 10, 2008 #1
    Two circles of radius 3√2 are tangent to the the graph y^2 =4x at (1,2). Find the equation of each circle.

    I have found the derivative of the graph, which is 1/√x. I know that the equation of the circle is X^2 + Y^2=r^2 where r is the radius so the equation of each circle is X^2 + Y^2=18. That is all I can get. Please help!
  2. jcsd
  3. Jun 10, 2008 #2


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    Okay, what is that at (1, 2)?

    No, you don't know that. That is the equation of a circle with center at (0,0) and one thing you can be pretty sure of is that these circles do NOT have center at (0,0)!

    If you have found the derivative at (1, 2) then you know the slope of the tangent line. What is the slope of, and what is the equation of, the line through (1, 2) perpendicular to the tangent line? I ask that because a tangent to a circle is always perpendicular to a radius. The centers of your two tangent circles must lie on the line through (1, 2) perpendicular to the tangent line to the curve, at distance [itex]3\sqrt{2}[/itex].
  4. Jun 10, 2008 #3
    okay so i have made progress...

    I plugged in x=1 into the derivative so the slope of my tangent is 1. Then I found the equation of the tangent which is y=x + 1. I now know that the slope of my circles is -1 and I know that
    (x-a)^2 + (y-b)^2= 18

    Could you help me with the next step??? In other words, how do I find a point with that slope and a distance or radius of 3sqrt2?
  5. Jun 10, 2008 #4
    The slope of the circles is not -1; this is the slope of the line perpendicular to the tangent.

    Once you've found the equation of this perpendicular line...

    The cool thing is that the slope of the perpendicular line is -1 so the change in the x and change in the y are the same. Hint: If you draw a triangle you will see that it is a special triangle. From here, solve for the side(s) of that triangle.

    Once you've done that, just simply add/subtract to (1,2) and you will get the centers of both circles. Finally, solve for the equation of the circles.
  6. Jun 11, 2008 #5
    i know that my equation of the line perpendicular to the tangent is y=3-x. From that, do I use my distance formula to get the centre of the circle... before it said something about special triangles- could you elaborate?
  7. Jun 11, 2008 #6


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    The centers of your two circles, (a, b), are ON the line y= 3- x, so you know b= 3- a. Further, they are distance [itex]3\sqrt{2}[/itex] from (1, 2). The distance from (a,b) to (1, 2) is [itex]\sqrt{(a-1)^2+ (b-2)^2}= \sqrt{(a-1)^2+ (3-a-2)^2}[/itex]
    Set that equal to [itex]3\sqrt{2}[/itex] and solve for a.
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