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    The circle with equation x² - 6x+y² - 4y = 0 crosses the y-axis at the origin and the point A.

    The first question asks me to find the co-ordinates of A. Now I feel this question is meant to be dead easy but I just can't see where to start with it.

    I rearranged it to (x-3)² + (y-2)² = 13

    But I still don't think that would help me.
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    When something crosses the y axis, what does it tell you about one of the co-ordinates? Think of any co-ordinates across the y axis. They all have something in common.
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    (Original post by SeanFM)
    When something crosses the y axis, what does it tell you about one of the co-ordinates? Think of any co-ordinates across the y axis. They all have something in common.
    That x = 0
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    Bingo. Now find the y co-ordinate and you've found the co-ordinates of A. :cool:
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    (Original post by SeanFM)
    Bingo. Now find the y co-ordinate and you've found the co-ordinates of A. :cool:
    Sorry for being really dumb but i'm still slightly confused.

    I'm confused am I not looking for a completely new point? If something passes through the origin then it's coordinates would be (0,0)

    Sorry again for being really confused :rolleyes:
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    You know that x=0 at A so stick zero in for x in the equation. That'll give you a nice quadratic in y.

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    In some sense yes, you're looking for a new point. There are two points on the circle where x is 0. One of them is the origin as you've said, and the other one is also on the y axis but has a different y co-ordinate.

    And above ^^

    Oh, and some advice for circle questions, sketch them if you can. In this case a radius of sqrt 13 is a bit icky though. But if you called it 3.5 you'd probably be able to draw a helpful circle.
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    (Original post by SeanFM)
    In some sense yes, you're looking for a new point. There are two points on the circle where x is 0. One of them is the origin as you've said, and the other one is also on the y axis but has a different y co-ordinate.

    And above ^^
    Oh I see now, So y² - 4y = 0

    y(y-4) = 0

    Therefore my coordinates would be (0,4) right?
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    Yeah, and just as a sanity check y=0 (the origin) also satisfies that quadratic. You've found A.
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    (Original post by SeanFM)
    Yeah, and just as a sanity check y=0 (the origin) also satisfies that quadratic. You've found A.
    Thanks a lot guys, I did make that a lot more difficult than it needed to be!

    Thanks again!
 
 
 
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