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Math Help - Help with circle geometry!

  1. #1
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    Help with circle geometry!

    I have a series of questions that I did. They lead up to the last question I can't solve. Could you check my math and help me with the last question? Thanks!


    a) Let a be the point (2,3). Compute the distance from Origin 0 to A


    answer: a^2 + b^2 = c^2
    3^2 + 2^2 = c^2
    c=sqrt(13)

    b) find the equation of circle C passing through point A

    x^2 + y^2 = 13 -> since the center is (0,0), right?

    c) find the equation of line D tangent to the circle C at point A
    since the equation for line OA is y=3/2 x + 0 , I can use the negative inverse of the slope to get the slope of the tangent, right?
    so I used y=-2/3 x +b and input the coordinates (2,3) to get b


    3 = (-2/3)(2) + b

    b = 4 , therefore the equation of the tangent is y=-2/3 x + 4 right?

    d) line D meet Ox at point B. Find the coordinates of B.
    This is where I'm a little confused. Does "Ox" mean the x-axis? That's what I went on so I just used the previous line equation and set y to zero:

    0= -2/3 x + b
    x = 6 therefore line D meets Ox when x=6 coordinates (6,0)

    e) compute distance of AB
    A(2,3) B(6,0) I used Pythagoreans theorem and got a distance of 5

    f) find the equation of the circle C' with center B and passing through A

    since the circle equation is (x-h)^2 + (y-k)^2 = R^2 I just input everything I knew so far and got: (x-6)^2 + y^2 = 25

    is all that correct? I know it's a lot but I appreciate the help!

    The last question is:

    g) find the coordinates of the intersection points of C and C'
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  2. #2
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    Quote Originally Posted by bloodyfinger View Post
    I have a series of questions that I did. They lead up to the last question I can't solve. Could you check my math and help me with the last question? Thanks!


    a) Let a be the point (2,3). Compute the distance from Origin 0 to A


    answer: a^2 + b^2 = c^2
    3^2 + 2^2 = c^2
    c=sqrt(13)

    b) find the equation of circle C passing through point A

    assuming C is centred at the origin. Has the question specified this ?

    x^2 + y^2 = 13 -> since the center is (0,0), right?

    c) find the equation of line D tangent to the circle C at point A
    since the equation for line OA is y=3/2 x + 0 , I can use the negative inverse of the slope to get the slope of the tangent, right?
    so I used y=-2/3 x +b and input the coordinates (2,3) to get b


    3 = (-2/3)(2) + b

    b = 4 , therefore the equation of the tangent is y=-2/3 x + 4 right?

    You have an error there, -4/3 +13/3=3

    d) line D meet Ox at point B. Find the coordinates of B.
    This is where I'm a little confused. Does "Ox" mean the x-axis? That's what I went on so I just used the previous line equation and set y to zero:

    0= -2/3 x + b
    x = 6 therefore line D meets Ox when x=6 coordinates (6,0)

    e) compute distance of AB
    A(2,3) B(6,0) I used Pythagoreans theorem and got a distance of 5

    f) find the equation of the circle C' with center B and passing through A

    since the circle equation is (x-h)^2 + (y-k)^2 = R^2 I just input everything I knew so far and got: (x-6)^2 + y^2 = 25

    is all that correct? I know it's a lot but I appreciate the help!

    The last question is:

    g) find the coordinates of the intersection points of C and C'
    .
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    .
    Does this make sense for c) ?

    3 = (-2/3)(2) + b

    b = 4/3 + 3
    b = 13/3
    so basically the y intercept is 13/3

    Also, continuing on,

    d) line D meets Ox when x=6.5 according to my math

    e) the line from point A(2,3) to B(6.5,0) is now \sqrt{117}/2

    f) Circle C' with center (6.5,0) passing through point (2,3) would have a radius of \sqrt{117}/2 and the equation would be:  (x-13/2)^2 + y^2 = 117/4

    right? this just all seems wrong to me though... sorry if this gets confusing
    Last edited by bloodyfinger; October 14th 2010 at 01:39 PM.
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  4. #4
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    Quote Originally Posted by bloodyfinger View Post
    Does this make sense for c) ?

    3 = (-2/3)(2) + b

    b = 4/3 + 3
    b = 13/3
    so basically the y intercept is 13/3

    Also, continuing on,

    d) line D meets Ox when x=6.5 according to my math

    e) the line from point A(2,3) to B(6.5,0) is now \sqrt{117}/2

    f) Circle C' with center (6.5,0) passing through point (2,3) would have a radius of \sqrt{117}/2 and the equation would be:  (x-13/2)^2 + y^2 = 117/4

    right? this just all seems wrong to me though... sorry if this gets confusing
    Yes, that's all fine.
    You only have part (g) left.
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Yes, that's all fine.
    You only have part (g) left.
    Thanks so much!! the answer for e just seems like such an odd number that it shouldn't really be the right answer. Oh well!

    Could you give me a hit as to how I can find the intersection points between two circles? I haven't a clue...
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  6. #6
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    Quote Originally Posted by bloodyfinger View Post
    Thanks so much!! the answer for e just seems like such an odd number that it shouldn't really be the right answer. Oh well!

    Could you give me a hint as to how I can find the intersection points between the two circles? I haven't a clue...
    Your work has been quite good.
    A clue for (g) is.....

    Point A is on both circles.
    Both circles have their centres on the x axis.
    Therefore the common chord is vertical, parallel to the y-axis and passes through the points of intersection.
    The x-axis serves as an axis of symmetry.

    What then do you reckon the two points of intersection are ?
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    Your work has been quite good.
    A clue for (g) is.....

    Point A is on both circles.
    Both circles have their centres on the x axis.
    Therefore the common chord is vertical, parallel to the y-axis and passes through the points of intersection.
    The x-axis serves as an axis of symmetry.

    What then do you reckon the two points of intersection are ?
    Ah of course, good point. So the other point would just be mirrored on the x-axis. 2,-3 right? Is there an actual formula to find the intersection points of two circles instead of just using mental gymnastics?

    Also, the one thing that worries me about this question is in part d) when he say "Line D meets Ox at point B". I'm just second guessing what Ox means. I interpreted it to mean that he wants me to find the X intercept of Line D. Is there any other way to interpret that?

    Thanks again!!!
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  8. #8
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    Quote Originally Posted by bloodyfinger View Post
    Ah of course, good point. So the other point would just be mirrored on the x-axis. 2,-3 right? Is there an actual formula to find the intersection points of two circles instead of just using mental gymnastics?

    Yes.... see below

    Also, the one thing that worries me about this question is in part d) when he say "Line D meets Ox at point B". I'm just second guessing what Ox means. I interpreted it to mean that he wants me to find the X intercept of Line D. Is there any other way to interpret that?

    I'm not sure! Hopefully not as I can only interpret it in the same way.

    Thanks again!!!
    To find the points of intersection of circles in general (2 at most unless they are the same circle)...

    The points of intersection are (p,q).

    x^2+y^2=13 refers to all the points (x,y) that are \sqrt{13} units from the origin.

    (x-6.5)^2+y^2=\frac{117}{4} refers to all points that are \sqrt{\frac{117}{4}} units from (6.5, 0).

    To find the points (p,q).... these points are on both circles.
    Hence we can write one co-ordinate in terms of the other, using one circle equation.
    Then substitute that into the 2nd equation to get a resulting equation in one co-ordinate.
    The solutions of that equation are the values of that co-ordinate.
    Then the 2nd co-ordinate is easily found.

    p^2+q^2=13\Rightarrow\ q^2=13-p^2

    \left(p-6.5\right)^2+13-p^2=\frac{117}{4}

    Solving for p

    p^2-13p+(6.5)^2-p^2+13-\frac{117}{4}=0

    13p=26

    p=2\Rightarrow\ 2^2+q^2=13\Rightarrow\ q^2=9\Rightarrow\ q=\pm3
    Last edited by Archie Meade; October 15th 2010 at 06:24 AM.
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