C2 circle question.

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I_Like_Maths
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#1
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#1
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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Kevin De Bruyne
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#2
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#2
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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I_Like_Maths
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#3
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#3
(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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Kevin De Bruyne
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#4
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#4
Bingo. Now find the y co-ordinate and you've found the co-ordinates of A. :cool:
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I_Like_Maths
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#5
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#5
(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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weasdown
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#6
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#6
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.

Posted from TSR Mobile
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Kevin De Bruyne
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#7
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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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I_Like_Maths
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#8
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#8
(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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Kevin De Bruyne
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#9
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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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I_Like_Maths
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#10
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#10
(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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