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Find the equation of the circle

  1. Jun 28, 2004 #1
    A circle touches the line 2x + 3y +1 = 0 at the pt. (1,-1) and is orthogonal
    to the circle whose one pair of diametrically opposite end pts. are (3,0) and
    (1,-3).Find the equation of the circle.
  2. jcsd
  3. Jun 28, 2004 #2


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    You know the diameter of the second circle is [itex]\sqrt{(3-1)^2 + [0 - (-3)]^2} = \sqrt{13}[/itex]. Therefore, the radius, [itex]r = \sqrt{13}/2[/itex], and [itex]r^2 = 13/4[/itex]

    Now the second circle can be represented by the equation:

    [tex](x-a)^2 + (y-b)^2 = r^2[/tex]

    [tex](x - a)^2 + (y - b)^2 = 13/4[/tex]

    You can make two equations with two unknowns, and solve for a and b:

    [tex](1):\ \ (3-a)^2 + b^2 = 13/4[/tex]

    [tex](2):\ \ (1-a)^2 + (3+b)^2 = 13/4[/tex]

    [tex]a = 2,\ b = -1.5[/tex]

    This should help you get started.

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  4. Jun 28, 2004 #3


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    Now, let (m,n) be the center of the circle we are solving for. We know:

    [tex](1-m)^2 + (1+n)^2 = R^2[/tex]

    where R is the radius of the circle we're solving for. We also know that there must be points on the other circle, [itex](x_1,y_1)[/itex] and [itex](x_2,y_2)[/tex] such that:

    [tex](x_1 - 2,y_1 + 1.5) \cdot (x_1 - m, y_1 - n) = 0[/tex]

    [tex](x_2 - 2,y_2 + 1.5) \cdot (x_2 - m, y_2 - n) = 0[/tex]

    We also have:

    [tex](x_1 - 2)^2 + (y_1 + 1.5)^2 = 13/4[/tex]

    [tex](x_2 - 2)^2 + (y_2 + 1.5)^2 = 13/4[/tex]


    [tex]\frac{1-m}{-1-n} = 3/2[/tex]


    [tex](x_1-m)^2 + (y_1-n)^2 = R^2[/tex]

    [tex](x_2-m)^2 + (y_2-n)^2 = R^2[/tex]

    8 equations, 7 unknowns. Looks ugly, but possible.
  5. Jun 29, 2004 #4


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    Okay, perhaps someone can explain to me why this approach didn't work.

    I started out with the three pieces of information that we have to begin with:

    1) The equation of the circle is [tex](x - a)^2 + (y - b)^2 = 13/4[/tex]

    2) The equation of the tangent line is [tex]2x + 3x + 1 = 0[/tex]

    3) The line is tangent to the circle at the point [tex](1,-1)[/tex]

    In other words, [tex]\frac{dy}{dx}[/tex] at [tex](1,-1)[/tex] is [tex]-2/3[/tex]

    So I differentiated the equation of the circle to obtain the equation for [tex]\frac{dy}{dx}[/tex]

    [tex]\frac{d}{dx}(x - a)^2 + \frac{d}{dx}(y - b)^2 = \frac{d}{dx}13/4[/tex]

    If [tex]u = x - a[/tex] and [tex]v = y - b[/tex],

    then [tex]u\prime = 1[/tex] and [tex]v\prime = 1[/tex]

    I rewrote the equation as

    [tex]\frac{d}{dx}u^2 + \frac{d}{dx}v^2 = \frac{d}{dx}13/4[/tex]

    [tex]2uu\prime + 2vv\prime = 0[/tex]

    [tex]2(x - a) + 2(y - b)\frac{dy}{dx} = 0[/tex]

    Solving for [tex]\frac{dy}{dx}[/tex], I got

    [tex]2(y - b)\frac{dy}{dx} = -2(x - a)[/tex]

    [tex]\frac{dy}{dx} = \frac{-2(x - a)}{2(y - b)} = -\frac{(x - a)}{(y - b)}[/tex]

    I then substituted in the known values for [tex](x,y)[/tex] and [tex]\frac{dy}{dx}[/tex],

    [tex](1,-1)[/tex] and [tex]-2/3[/tex]

    [tex]-2/3 = -\frac{(1 - a)}{(1 - b)}[/tex]

    The negatives cancel and we get

    [tex]1 - a = 2[/tex] and [tex]-1 - b = 3[/tex]

    [tex]a = -1[/tex] and [tex]b = -4[/tex]

    Substituting these values back into the equation of the circle, we get

    [tex](x + 1)^2 + (y + 4)^2 = 13/4[/tex]

    Now this equation is not correct, as it never touches the point [itex](1,-1)[/itex]. I'm sure I'm overlooking something very simple, but it's 1:30 AM here and I'm really tired. Can someone help me out?
    Last edited: Jun 29, 2004
  6. Jun 29, 2004 #5
    Yes, you are overlooking something very simple: the circle we're asked to find doesn't (necessarily) have a radius of sqrt(13/4).

    The circle with "diametrically opposite end pts. (3,0) and (1,-3)" has a radius of sqrt(13/4). Its equation is easy to find, since you know two points on it. AKG did it in his first post. This circle is not a tangent to 2x + 3y + 1 = 0.

    What we're looking for is a totally different circle (that IS a tangent to 2x + 3y + 1 = 0). The only connection this new circle has with the old circle, is that they are orthogonal to each other.

    *edit* That's not to say that you can't find a circle with a radius of sqrt(13/4) that is tangent to 2x + 3y + 1 = 0 at (1, -1). Let the circle's equation be (x - a)^2 + (y - b)^2 = 13/4. Work out its intersection with the tangent line. The discriminant must be zero. You now have an equation relating a and b. But since you also know that (1, -1) must lie on the circle, you have another equation with a and b. Two equations, two variables, should be solvable. In fact, if you slog through the calculations (or enlist a computer to do it for you ;)), you'll find that the equation is x^2 + (y + 5/2)^2 = 13/4 or (x - 2)^2 + (y - 1/2)^2 = 13/4.

    You can apply the same technique (letting the circle's equation be (x - a)^2 + (y - b)^2 = r^2, though) to solve the original problem, only you have to apply the condition given on Mathworld... Very un-pretty though.

    And now I see the mistake you did. If you have this equation:

    [tex]-2/3 = -\frac{(1 - a)}{(1 - b)}[/tex]

    1 - a is not necessarily equal to 2, and 1 - b is not necessarily equal to 3. Since [tex]\frac{2}{3} = \frac{4}{6}[/tex], is 2 = 4 and 3 = 6?
    Last edited: Jun 29, 2004
  7. Jun 29, 2004 #6


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    A lot of the calculations you did are not necessary: The circle such that (3, 0) and (1, -3) has radius [itex]\frac{\sqrt{13}}{2}[/itex] AND center at ((3+1)/2,(0-3)/2)=
    (2, -3/2) so THAT circle is (x-2)2+ (y+ 3/2)2= 13/4.

    The circle you want to find must be of the form (x-a)2+ (y-b)2= R2 (any circle is of that form). If it "touches" (is tangent to) the line
    2x+ 3y+ 1= 0 then it must be tangent at (x,y) such that 2x+ 3y+ 1= 0 and
    x2+ y2= R2. That's two equations for 3 unknowns.

    In order to be perpendicular to the circle (x-2)2+ (y+ 3/2)2= 13/4, The must be two points (x,y) satisfying the equations of both circles and such that the derivatives y'= (y- 3/2)/(x- 2)= (y-b)/x-a). You need to solve all those equations for a and b.
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