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Homework Help: Circle Exercise Question

  1. Mar 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Equation: x^2+y^2-6x-2y+8=0 Find the center and the radius.
    (Help) : Find the equation of the tangent to the circle above that passes through the beginning of axis O (0,0)

    3. The attempt at a solution
    I found the center and radius and i believe the values are : C (3,1) and R =root(2) but can you explain me the second one

    Sorry for my english.

    I hope you'll understand it.

    Thank you.
  2. jcsd
  3. Mar 9, 2017 #2
    Your center and radius check out:
    (x-3)^2 + (y-1)^2 -2 = x^2-6x+9+y^2-2y+1 -2 = x^2 + y^2 - 6x - 2y +10 - 2

    Since your tangent line crosses through (0,0), it will be of the form x=my, where m is the slope. It will be perpendicular to your circle, so the formula for a line crossing through the tangent point and the center of the circle will be (x-3) = (-1/m)(y-1).

    Now you need to find a slope "m" that will create an intersection between those equations that is on the circle: (x^2 + y^2 - 6x - 2y +8 = 0).

    If you draw a picture, you will notice that there are two solutions. In both cases, a right triangle can be formed between (0,0), (3,1), and the tangent point. In both cases, the length of the hypotenuse will be sqrt(3^2+1^2) and the length of the leg that extends from the center of the circle will be sqrt(2). So the distance of the tangent points to the origin (0,0) can be easily computed.
    So both tangent points will be on a circle centered at (0,0) with that radius. I'll let you set up that equation.
  4. Mar 9, 2017 #3


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    How did you find the center and radius?
  5. Mar 9, 2017 #4
    I'm assuming that you only want an answer from the OP on this.
  6. Mar 9, 2017 #5


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    Staff Emeritus
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    Yeah, I misread what the OP was asking. Thought s/he had obtained the center and radius but was asking why the radius was sqrt(2).
  7. Mar 9, 2017 #6
    The equation for the tangent line is going to be y = kx, where k needs to be determined. You are looking for the intersection of y = kx with your circle. If you substitute y = kx into the equation for your circle, you get a quadratic equation, with the roots being the two points of intersection. If the line is tangent, there will only be one point of intersection (double root). This will be when the discrimanent is equal to zero.
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