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Math Help - Circle on Plane

  1. #1
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    Circle on Plane

    Hey, me again.

    The centres A,B, and C of the three circles are collinear. (A is smallest circle, B is biggest, C is 2nd biggest)



    The equation of circle A is: (x + 12) + (y + 15) = 25
    The equation of circle C is: (x - 24) + (y - 12) = 100

    Find the equation of circle B.


    ------

    I done the easy part of finding the radius of Circle B (15). The bit i'm stuck at is finding the centre of circle B, which from the answer is (4,-3) . Thanks


    The answer is: x + y - 8x + 4y + 200 = 0
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    If (h,k) is the center of B then the distance from (-12,-15) to (h,k) is 20.
    The slope (k+15)/(h+12) must be 3/4.
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  3. #3
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    Quote Originally Posted by Plato View Post
    If (h,k) is the center of B then the distance from (-12,-15) to (h,k) is 20.
    The slope (k+15)/(h+12) must be 3/4.
    I don't quite understand the part in bold, unless your talking about the equation of the line.
    A(-12, -15) C(24, 12) m = 3/4
    4y = 3x - 24

    Could you say what the next step is, in finding the centre of B.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by r_maths View Post
    Hey, me again.

    The centres A,B, and C of the three circles are collinear. (A is smallest circle, B is biggest, C is 2nd biggest)



    The equation of circle A is: (x + 12) + (y + 15) = 25
    The equation of circle C is: (x - 24) + (y - 12) = 100

    Find the equation of circle B.


    ------

    I done the easy part of finding the radius of Circle B (15). The bit i'm stuck at is finding the centre of circle B, which from the answer is (4,-3) . Thanks


    The answer is: x + y - 8x + 4y + 200 = 0
    There are actually many ways to do this problem, this is the easiest way i can see at the moment, but then i never thought about it very long.

    The radius of A is 5, the radius of C is 10 (do you see that?)
    the center of A is (-12,-15), the center of C is (24,12) (do you see that?)

    what is the distance between the two centers?

    d = sqrt[(24 + 12)^2 + (12 + 15)^2] = sqrt(1296 + 729) = 45

    since the radius of A is 5 and the radius of C is 10, then the diameter of B must be:

    45 - 5 - 10 = 30
    so the radius of B is 15.

    now the slope of the line connecting the center of A and the center of B is:
    m = (y2 - y1)/(x2 - x1) = (12 + 15)/(24 + 12) = 27/36 = 3/4

    so notice that if we move 4 units horizontally from the center of A and 3 units up as the slope suggests, we end up with a 3-4-5 triangle. the hypotenuse of this triangle will be 5, which is the exact radius of A. the distance between the center of A and the center of B is 20. so the hypotenuse of the triangle that connects the center of A to the center of B is 20. that is we multiply 5 by 4. since this will be a similar triangle to the first, we can multiply the other 2 numbers by 4. that is we move 4*4 units across and 3*4 units up. that is 16 units across and 12 units up. so the center of B will be given by:
    where (x,y) is the center of A

    (or we could use: 20/y = 5/3 and 20/x = 5/4 to find the x and y units we move from the center of A)

    (x + 16, y + 12) = (-12 + 16,-15 + 12) = (4,-3).

    Now we know the radius of B and it's center.

    so the equation of B is:

    (x - 4)^2 + (y + 3)^2 = 15^2
    => x^2 - 8x + 16 + y^2 + 6y + 9 = 225
    => x^2 - 8x + y^2 + 6y - 200 = 0




    another way: we can find the center of B by finding the point (x,y) that is 20 units from the center of A and at the same time is 25 units from the center of C

    so we have:

    25 = sqrt[(x - 24)^2 + (y - 12)^2] ..................(1)
    20 = sqrt[(x + 12 )^2 + (y + 15 )^2] ...............(2)

    solving this system should give you the center coordinates, but that was just too much work

    also, you could find the equation of the line:

    what is the line connecting these three points?
    let's find the slope first:
    m = (y2 - y1)/(x2 - x1) = (12 + 15)/(24 + 12) = 27/36 = 3/4

    using the point slope form with m = 3/4 and (x1,y1) = (24,12), we get:
    y - y1 = m(x - x1)
    => y - 12 = (3/4)(x - 24)
    => y = (3/4)x - 18 + 12
    => y = (3/4)x - 6

    and use the distance formula with the relevant data to get the points for the center of B, using the equation of the line as a constraint, again, too much work. simialr triangles seems easiest.

    i can attach a diagram if you want to see what similar triangles i am talking about.

    did you get my solution?
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  5. #5
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    Quote Originally Posted by r_maths View Post
    about the equation of the line.
    A(-12, -15) C(24, 12) m = 3/4
    4y = 3x - 24
    (h+12)^2 + (y+15)^2 = 20^2 and 4(k+15)=3(h+12).
    Now solve for (h,k).
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    did you get my solution?
    Yep. I know what sim. triangle your talking about, Thanks for your time.
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