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Show that the centers of the circles passing through the points

  1. Dec 25, 2006 #1
    1. The problem statement, all variables and given/known data
    Show that the centers of the circles passing through the points (3,2) and (6,3) are located on the line 3x+y=16.

    Two of these circles touch the line x+2y=2. Find the equation of both these circles.

    2. Relevant equations

    The general equation of curves (circles, parabola, ellipse, hyperbola)

    3. The attempt at a solution
    The proving section is fine to me, I merely need to find the perpendicular bisector.

    I approached the question by trying to search for the centers of the circles. I expected myself to come to a linear equation and a quadratic equation (since there is two circles) One equation is clear : 3x+y=16

    But I can't find the other one. Should I deal with the gradients? Or the radius of the circles?
  2. jcsd
  3. Dec 25, 2006 #2


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    Let (x,y) satisfy x+2y=2, and let (x',y') satisfy 3x'+y'=16. Now you want a circle centered at (x',y'), passing through (3,2) and (6,3), and touching the line x+2y=2 at (x,y). You've got four unknowns, so you want four equations. You already have two. You can get two more because you know:

    a) the distance from (x,y) to (x',y') is the same as the distance from (3,2) to (x',y')
    b) the line segment from (x,y) to (x',y') must be perpendicular to the line 2x+y=2.

    Note that although you also know that the distance from (x,y) to (x',y') must be the same as the distance from (6,3) to (x',y'), the equation you get from this fact doesn't help you solve anything, because it immediately follows from the equations 3x'+y'=16 and the equation you get from a) above, and thus it would just be redundant information.
  4. Jan 2, 2007 #3
    All of the circles have equation (x - a)^2 + (y - b)^2 = r^2 where b and a > 0, r is the radius, and (a,b) is the center. Substiting the points through which the circles pass gives:

    (3 - a)^2 + (2 - b)^2 = r^2
    (6 - a)^2 + (3 - b)^2 = r^2

    Therefore (3 - a)^2 + (2 - b)^2 = (6 - a)^2 + (3 - b)^2

    Expanding this leads to solutions b = 16 - 3a and a = (16 - b)/3

    If (a,b) were to lie on the line 3x + y = 16, its coordinates could be expressed as (x, 16 - 3x) for any x. When we say for "any x," that includes a, so therefore (a, 16 - 3a) is a point, which is equivalent to (a, b). Therefore the center of these circles lie on the line 3x + y = 16

    You're then told that 2 of the circles touch the line x + 2y = 2, meaning both circles only have one point of contact, (x, 1 - x/2)

    You can proceed using the discriminant of the intersection equation or substituting values of a and b.
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