Coordinate Geometry and circle

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  1. krisshP's Avatar
    1 04-07-2012  11:00
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    Coordinate Geometry and circle
    I need help on part c of this question below:

    a. Write down the coordinates of the mid-point M of the line joning A(0, 2) and B(4, 8).

    I understand this part and got the correct answer of (2, 5).

    b. Show that the line 3y+2x-19=0 passes through M and is perpendicular to AB.

    I made 3y+2x-19=0 be in the form of y=mx+c by making it become the following:

    y=-\frac{2}{3}x+\frac{19}{3}

    Therefore I ended up proving that when x=2, y=5.
    I got 1.5 as the gradient of line AB and proved the line 3y+2x-19=0 is perpendicular to line AB.

    Part c, which is below is difficult for me right now, is below:

    c. Calculate the coordinates of the centre of the circle which passes through A, B and the origin O.

    I'm not even sure how to start part c because it seems quite complicated for me currently.:confused::confused:

    Thanks a lot for any help provided.
    Last edited by krisshP; 04-07-2012 at 11:04.
  2. aznkid66's Avatar
    2 04-07-2012  11:05
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    Re: Coordinate Geometry and circle
    Circle: locus of points equidistant to the [circle's] center.

    Therefore, the distances (hint: distance formula) from any point on the circle (hint: A, B, and (0,0) ) and the center (hint: coordinate with the two variables you're trying to find) are equal to eachother.
  3. krisshP's Avatar
    3 04-07-2012  11:11
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    Re: Coordinate Geometry and circle
    (Original post by aznkid66)
    Circle: locus of points equidistant to the [circle's] center.

    Therefore, the distances (hint: distance formula) from any point on the circle (hint: A, B, and (0,0) ) and the center (hint: coordinate with the two variables you're trying to find) are equal to eachother.
    So are you really suggesting that I should draw a graph with line AB and then find the point where my compasses stand on and then find the coordinates????

    How would I do this in a C1 exam? You are not allowed compasses in a C1 exam.:confused:
  4. Plato's Trousers's Avatar
    4 04-07-2012  11:31
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    So are you really suggesting that I should draw a graph with line AB and then find the point where my compasses stand on and then find the coordinates????

    How would I do this in a C1 exam? You are not allowed compasses in a C1 exam.:confused:
    The equation of a circle with centre (a,b) and radius r is
    (x-a)^2 + (y-b)^2 = r^2

    The distance from each point (A, B and the origin) to the centre (C) will be the same (r) since they all lie on the circle. Write out three distance equations (for AC, BC and OC) and solve

    d_{AC}=\sqrt{(0-a)^2 +(2-b)^2}=r

    d_{BC}=\sqrt{(4-a)^2 +(8-b)^2}=r

    and so on
    Last edited by Plato's Trousers; 04-07-2012 at 11:41.
  5. aznkid66's Avatar
    5 04-07-2012  11:51
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    So are you really suggesting that I should draw a graph with line AB and then find the point where my compasses stand on and then find the coordinates????

    How would I do this in a C1 exam? You are not allowed compasses in a C1 exam.:confused:
    What? Did I mention drawing anything at all? All you need to do is plug in values into the distance formula because that's what the definition of a circle tells you to do. I even bolded the single and only word of any importance in the seven-word definition.

    Distance formula for distance between (x_1,y_1) and (x_2,y_2):

    \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}

    Let 'C' be the center of the circle and 'O' be the origin. Because A, B, and O lie on the same circle centered at C, CA=CB=CO. This gives you three equations: CA=CB, CA=CO, CB=CO with two variables (the coordinates of the center).
    Last edited by aznkid66; 04-07-2012 at 11:52.
  6. Plato's Trousers's Avatar
    6 04-07-2012  11:59
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    So are you really suggesting that I should draw a graph with line AB and then find the point where my compasses stand on and then find the coordinates????

    How would I do this in a C1 exam? You are not allowed compasses in a C1 exam.:confused:

    (Original post by aznkid66)
    What? Did I mention drawing anything at all? All you need to do is plug in values into the distance formula because that's what the definition of a circle tells you to do. I even bolded the single and only word of any importance in the seven-word definition.

    Distance formula for distance between (x_1,y_1) and (x_2,y_2):

    \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}

    Let 'C' be the center of the circle and 'O' be the origin. Because A, B, and O lie on the same circle centered at C, CA=CB=CO. This gives you three equations: CA=CB, CA=CO, CB=CO with two variables (the coordinates of the center).
    krisshP - in case you'e confused, my equations and aznkid66's are the same, it's just that I have done part of the work for you. There are really only two steps to this;

    1 the distances AC, BC, OC are all equal
    2 the distances are given by the distance formula
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  7. krisshP's Avatar
    7 04-07-2012  12:05
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    Re: Coordinate Geometry and circle
    For the distance of the line AC:
    \sqrt{(0-x)^2+(2-y)^2}=r

    \sqrt{x^2+y^2-4y+4}=r

    Shall I now square both sides or what??:confused:
  8. aznkid66's Avatar
    8 04-07-2012  12:10
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    Re: Coordinate Geometry and circle
    (Yes.)

    The last distance, d_{OC} is necessary. This is because, to solve for three variables (x, y, and r) you need 3 equations.

    To solve for two variables (you don't need to find r), you only need 2 equations with two variables (x and y but not r), and you can eliminate the r's in the equations by substituting r=r with the distances (which are in terms of x and y).
    Last edited by aznkid66; 04-07-2012 at 12:13.
  9. BabyMaths's Avatar
    9 04-07-2012  12:10
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    Re: Coordinate Geometry and circle
    You have (0,0) and (0,2) on the circle so the y coordinate of the centre is 1.

    You have the equation of the bisector of AB. Sub in y=1 to get the x coordinate of the centre.

    This is much simpler than some of the advice you have been given.
  10. blueray's Avatar
    10 04-07-2012  12:13
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    For the distance of the line AC:
    \sqrt{(0-x)^2+(2-y)^2}=r

    \sqrt{x^2+y^2-4y+4}=r

    Shall I now square both sides or what??:confused:
    AB is (2,5)

    The equation of a circle with center (a,b) and radius r is

    (x-a)^2 + (y-b)^2 = r^2

    So (x-2)^2 + (y-5)^2 = r^2


    I haven't done this for ages, am I right?
    Last edited by blueray; 04-07-2012 at 12:19.
  11. krisshP's Avatar
    11 04-07-2012  12:13
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    Re: Coordinate Geometry and circle
    (Original post by BabyMaths)
    You have (0,0) and (0,2) on the circle so the y coordinate of the centre is 1.
    Why? is it because (0,1) is the midpoint?
  12. krisshP's Avatar
    12 04-07-2012  12:14
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    Re: Coordinate Geometry and circle
    (Original post by blueray)
    A(0, 2) and B(4, 8).
    x1,y1 x2,y2


    Plug these values into the forumla.

    \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}
    Doesn't that end up providing the distance of the line AB, which I think is a chord?
  13. aznkid66's Avatar
    13 04-07-2012  12:15
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    Re: Coordinate Geometry and circle
    (Original post by BabyMaths)
    You have (0,0) and (0,2) on the circle so the y coordinate of the centre is 1.

    You have the equation of the bisector of AB. Sub in y=1 to get the x coordinate of the centre.

    This is much simpler than some of the advice you have been given.
    Silly BabyMaths, you can't draw a circle with only two points...
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  14. BabyMaths's Avatar
    14 04-07-2012  12:15
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    Why? is it because (0,1) is the midpoint?
    You have a chord (0,0) to (0,2).

    The equation of the bisector of this chord is y=1.

    The centre of a circle lies on the bisectors of chords.
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  15. blueray's Avatar
    15 04-07-2012  12:16
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    Doesn't that end up providing the distance of the line AB, which I think is a chord?
    I will edit it sorry.
  16. krisshP's Avatar
    16 04-07-2012  12:17
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    Re: Coordinate Geometry and circle
    (Original post by blueray)
    A(0, 2) and B(4, 8).
    ---x1,y1------ x2,y2

    see how one is x1 and y1 and the other is x2 and y2?


    Plug these values into the forumla.

    \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}
    Line AB distance is \sqrt{52}. How can I use this to find the coordinates of the centre of the circle????
  17. krisshP's Avatar
    17 04-07-2012  12:18
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    Re: Coordinate Geometry and circle
    (Original post by BabyMaths)
    You have a chord (0,0) to (0,2).

    The equation of the bisector of this chord is y=1.

    The centre of a circle lies on the bisectors of chords.
    Why can't line AB be a chord?
  18. blueray's Avatar
    18 04-07-2012  12:20
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    Line AB distance is \sqrt{52}. How can I use this to find the coordinates of the centre of the circle????
    Ignore that, look at it again
  19. aznkid66's Avatar
    19 04-07-2012  12:20
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    Re: Coordinate Geometry and circle
    To be fair, you weren't exactly clear with your process. Plus, I find canceling out squares to be easier than simultaneous with slope-point form, so I was expecting something "much simpler".
  20. BabyMaths's Avatar
    20 04-07-2012  12:21
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    Re: Coordinate Geometry and circle
    (Original post by krisshP)
    Why can't line AB be a chord?
    It is. You have the equation of the bisector of this chord too.

    It's pity you got such bad advice on this.

    Answer is (x-8)^2+ (y-1)^2=65
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