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Last minute P3 revision

Hey everyone!!!

I just need some help on circle theorems in P3. It is a thought i had and it is now beginning to annoy me!!!! :biggrin:

OK. You know when you have 2 circles that have the same radius and TOUCH, you say that the point where they touch is the midpoint of the ine connecting the circles together. Suppose that they cross and you need to find the point where they cross? Do you substitute one equation for the ohter? Cos that would be REALLY complicated - because splittling one of the equation in x's and y's seems really taxing. Also, suppose if two circles touch and they are of DIFFERENT radii, how do you find the point where they touch??? :confused:

Do you think that i am reading too much into circle theorems?

BTW: Good luck to everyone with there exams esp. maths!!! :biggrin:

Thanks everyone!!! :biggrin:
Reply 1
Silly Sally
Hey everyone!!!

I just need some help on circle theorems in P3. It is a thought i had and it is now beginning to annoy me!!!! :biggrin:

OK. You know when you have 2 circles that have the same radius and TOUCH, you say that the point where they touch is the midpoint of the ine connecting the circles together. Suppose that they cross and you need to find the point where they cross? Do you substitute one equation for the ohter? Cos that would be REALLY complicated - because splittling one of the equation in x's and y's seems really taxing. Also, suppose if two circles touch and they are of DIFFERENT radii, how do you find the point where they touch??? :confused:

Do you think that i am reading too much into circle theorems?

BTW: Good luck to everyone with there exams esp. maths!!! :biggrin:

Thanks everyone!!! :biggrin:
If you know they just touch and are tangents to each other, then find the centres and radii of both circles by looking at their formulae:

eg. (x-a)^2+(y-b)^2=r^2 so it would be centred on (a,b) and have radius r. You will then have two radii and two centres, so start at one end of this line and find the fraction of the line which is the first radius (so if C1 is centred on (a,b) and has radius r and C2 is centred on (c,d) and has radius s, then starting at a, the circles intersect at (a+r(c-a)/s,b+r(d-b)/s).
Sorry, that sounds far more complicated than it really is. If I've explained it poorly, post again and I'll take you through it better.
Reply 2
meepmeep
If you know they just touch and are tangents to each other, then find the centres and radii of both circles by looking at their formulae:

eg. (x-a)^2+(y-b)^2=r^2 so it would be centred on (a,b) and have radius r. You will then have two radii and two centres, so start at one end of this line and find the fraction of the line which is the first radius (so if C1 is centred on (a,b) and has radius r and C2 is centred on (c,d) and has radius s, then starting at a, the circles intersect at (a+r(c-a)/s,b+r(d-b)/s).
Sorry, that sounds far more complicated than it really is. If I've explained it poorly, post again and I'll take you through it better.


Do you mind going over it again - i got a little confused

Thanks :smile:
Reply 3
sounds like your doing the jun03 paper, for that u dont need the equations of the line


just get the co ordinates of the 2 circle centre's then find the midpoint of them using (x + x)/2 , (y + y)/2
Reply 4
Asha
sounds like your doing the jun03 paper, for that u dont need the equations of the line


just get the co ordinates of the 2 circle centre's then find the midpoint of them using (x + x)/2 , (y + y)/2


Actually i did sort of have that paper in mind - and when thinking about that circle theorem question on it - other questions came to mind - for example if the radii were different.
Reply 5
Silly Sally
Do you mind going over it again - i got a little confused

Thanks :smile:

Well, if you take two circles, C1 and C2 which have equations:

C1: (x-a)^2+(y-b)^2=r^2
C2: (x-c)^2+(y-d)^2=s^2

then the two circles have centres (a,b) and (c,d) and radii r and s respectively. We can now draw this on a diagram (attached below)

The red part shows radius r and the green part shows radius s, so the whole line has length r+s. The proportion of the line which is red is thus r/(r+s).

Now, all you need to do is find the change in vertical height (d-b) and the change in horizontal (c-a). The red part does r/(r+s) of these changes, so you can find out how far up and across the point is from the centre of C1. The just add this to the centre of C1 to get the point of intersection.


Is that better?
Reply 6
Asha
sounds like your doing the jun03 paper, for that u dont need the equations of the line


just get the co ordinates of the 2 circle centre's then find the midpoint of them using (x + x)/2 , (y + y)/2
Alternatively, that is a far better explanation of what I was trying to say.
Reply 7
meepmeep
Well, if you take two circles, C1 and C2 which have equations:

C1: (x-a)^2+(y-b)^2=r^2
C2: (x-c)^2+(y-d)^2=s^2

then the two circles have centres (a,b) and (c,d) and radii r and s respectively. We can now draw this on a diagram (attached below)

The red part shows radius r and the green part shows radius s, so the whole line has length r+s. The proportion of the line which is red is thus r/(r+s).

Now, all you need to do is find the change in vertical height (d-b) and the change in horizontal (c-a). The red part does r/(r+s) of these changes, so you can find out how far up and across the point is from the centre of C1. The just add this to the centre of C1 to get the point of intersection.


Is that better?



AHH!!! Thanks :biggrin:

Basically take proportions of length!!! Thanks - i understand it now!! :biggrin: :biggrin: :biggrin:

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