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# help Watch

so do you set y to 0 in the line equation to find the point the circle touches the x axis?
Well it's a bit more involved than that, but you need to understand visually on a graph what is happening before you can even attempt to do the algebra.

You do realise that touches, just means that the circle 'touches' the axes at one point, so there isn't a whole section of the circle over the axes.

Let me give you an example:

http://www.wolframalpha.com/input/?i...1%29%5E2+%3D+1

The above link, shows a circle 'touching' both axes.

The below link shows a circle which you wouldn't say is 'touching' both axes.

http://www.wolframalpha.com/input/?i...1%29%5E2+%3D+2
2. (Original post by Noble.)
Honestly, there's no point in me spoon feeding you all the way to the answer, it doesn't help you.

You know that the circle 'touches' the x-axis and the y-axis - do you know what this means?
yes it means that there is a solution where x=0 and y=0
3. (Original post by Noble.)
Well it's a bit more involved than that, but you need to understand visually on a graph what is happening before you can even attempt to do the algebra.

You do realise that touches, just means that the circle 'touches' the axes at one point, so there isn't a whole section of the circle over the axes.

Let me give you an example:

http://www.wolframalpha.com/input/?i...1%29%5E2+%3D+1

The above link, shows a circle 'touching' both axes.

The below link shows a circle which you wouldn't say is 'touching' both axes.

http://www.wolframalpha.com/input/?i...1%29%5E2+%3D+2
yea i get that, thanks

i think i am missing something very obvious
yea i get that, thanks

i think i am missing something very obvious
Right ok, so what is the condition on the centre point that must exist in order for a circle to touch both axes? Remember, it's a circle, so constant radius which implies that the distance between the point and the x-axis and the point and the y-axis are equal. So what must our centre point satisfy?
5. (Original post by Noble.)
Right ok, so what is the condition on the centre point that must exist in order for a circle to touch both axes? Remember, it's a circle, so constant radius which implies that the distance between the point and the x-axis and the point and the y-axis are equal. So what must our centre point satisfy?
it must satisfy the equation of the line?
so do you set y to 0 in the line equation to find the point the circle touches the x axis?
Consider what Noble said about the point (p,p) and the radius of a circle. For both axes to be tangents what has to be satisfied?
7. (Original post by joostan)
Consider what Noble said about the point (p,p) and the radius of a circle. For both axes to be tangents what has to be satisfied?
I have no idea

sorry to be blunt but it's the truth
it must satisfy the equation of the line?
Yes, obviously. But now think in terms of the axes, forget the equation y=3x-4 for now. If you want to draw a circle which touches the x-axis and the y-axis what must the centre point satisfy?

The set of points you're after, are the sets of points where you could take a pen, and if you drew a straight line from the point to the x-axis, and then drew a straight line to the y-axis they would be the same distance. What does this imply?
9. (Original post by Noble.)
Yes, obviously. But now think in terms of the axes, forget the equation y=3x-4 for now. If you want to draw a circle which touches the x-axis and the y-axis what must the centre point satisfy?

The set of points you're after, are the sets of points where you could took a pen, and if you drew a straight line from the point to the x-axis, and then drew a straight line to the y-axis they would be the same distance. What does this imply?
the radius is the same at all points of a circle?! eugh I dunno what else it could mean

sound so dumb
the radius is the same at all points of a circle?! eugh I dunno what else it could mean

sound so dumb
Ok, think of the point if you draw a line from this point to the x-axis, how long is the line? What about if you now draw a line from this point to the y-axis, how long is this line?

You want the points where if you draw a line from the point to the axes, they're the same length.
11. (Original post by Noble.)
Ok, think of the point if you draw a line from this point to the x-axis, how long is the line? What about if you now draw a line from this point to the y-axis, how long is this line?

You want the points where if you draw a line from the point to the axes, they're the same length.
(1,2)... 1 unit from y axis 2 units from x axis

(5,12)... 5 units from y axis , 12 units from x axis

and yea i know but how are we meant to find that point?
(1,2)... 1 unit from y axis 2 units from x axis

(5,12)... 5 units from y axis , 12 units from x axis

and yea i know but how are we meant to find that point?
It isn't a specific point, there are infinitely many points which satisfy what I'm asking you. You want the points where if it's p units from the y axis and q units from the x axis, p=q.... So what are the set of points?
13. (Original post by Noble.)
It isn't a specific point, there are infinitely many points which satisfy what I'm asking you. You want the points where if it's p units from the y axis and q units from the x axis, p=q.... So what are the set of points?
I don't know

I have no idea how to find these points

that's why I am so lost
I don't know

I have no idea how to find these points

that's why I am so lost
How far is the point from the axes? How about the point ? How about the point ? How about the point ? How about the point ? How about the point ? How about the point ?
15. (Original post by Noble.)
How far is the point from the axes? How about the point ? How about the point ? How about the point ? How about the point ? How about the point ? How about the point ?
they are all equal units form the axis..........................ju st as the centre coordinates of the circle are. I know this, but how do we find these coordinates? what method do I use to get these coordinates?
they are all equal points form the axis..........................ju st as the centre coordinates of the circle are. I know this, but how do we find these coordinates? what method do I use to get these coordinates?
If you knew these were the points, why didn't you say when I asked for the umpteenth time?

So, the circle must satisfy this and the points must also satisfy the equation y=3x-4

Ok, so now, what must the centre of the circle be?
17. (Original post by Noble.)
If you knew these were the points, why didn't you say when I asked for the umpteenth time?

So, the circle must satisfy this and the points must also satisfy the equation y=3x-4

Ok, so now, what must the centre of the circle be?

THIS part is my problem. This is the point I am not sure of ..

THIS part is my problem. This is the point I am not sure of ..
Please tell me you're now winding me up? I've pretty much spoonfed you the answer and you still don't get it. You're not reading what I'm writing, or clearly not fully understanding something. The point must satisfy the condition that it's of equal distance to both axes and it must satisfy y=3x-4.
19. (Original post by Noble.)
Please tell me you're now winding me up? I've pretty much spoonfed you the answer and you still don't get it. You're not reading what I'm writing, or clearly not fully understanding something. The point must satisfy the condition that it's of equal distance to both axes and it must satisfy y=3x-4.
No! I am not winding you up! dont worry

But yea, I know that... but where do I get these numbers?!! there are no coordinates in the question or anything?? Eugh, i give up on this, ill move onto the next question on the book...
No! I am not winding you up! dont worry

But yea, I know that... but where do I get these numbers?!! there are no coordinates in the question or anything?? Eugh, i give up on this, ill move onto the next question on the book...

For the circle to 'touch' each axis we must have that the centre point of the circle is of the form where

Since we also have that lies on the line we must have but we also know that , so gives the centre of the circle being now the radius is also obvious.

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