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# Coordinates Of Circle Center Further Maths GCSE watch

1. Absolutely no idea how to do this one...
2. (Original post by Retsek)

Absolutely no idea how to do this one...
Are you familiar with the factorised form of the equation of a circle?

I.e.
3. (Original post by Retsek)
Absolutely no idea how to do this one...

Now complete the squares on x and y inside the bracket.
4. (Original post by Retsek)

Absolutely no idea how to do this one...
Also this one, is it just trial and error?
5. (Original post by RDKGames)

Now complete the squares on x and y inside the bracket.
Thanks a lot! I got that far before getting myself confused
6. (Original post by Retsek)
Also this one, is it just trial and error?
Start visualising and drawing sketches.

if one of those lines are a tangent to the circle, then you know that one of those lines touch the circle at a single point. Bang all of them in to the circle equation and find out which touch the circle at a single point
7. (Original post by Retsek)
Thanks a lot! I got that far before getting myself confused
3(x^2+y^2-4x-3y)+1=0
3((x^2-4x)+(y^2-3y))+1=0

You know about completing the square right? As RDK said, just do that on the two inner brackets, and you'll get the answer (after some rearranging).
8. (Original post by Retsek)
Also this one, is it just trial and error?
(Original post by Naruke)
Start visualising and drawing sketches.

if one of those lines are a tangent to the circle, then you know that one of those lines touch the circle at a single point. Bang all of them in to the circle equation and find out which touch the circle at a single point
While this is rigorous and bound to give the correct answer, a quicker approach is to notice that the radius is 2 and centre at the origin - and so you're looking for a line which has a minimum distance from the origin as 2.

For example; the fourth equation fails this because the line crosses through and and since you have the line going through 2 axis points which are inside the circle, with minimum distance being less than 2 as a result.
9. (Original post by Darth_Narwhale)
3(x^2+y^2-4x-3y)+1=0
3((x^2-4x)+(y^2-3y))+1=0

You know about completing the square right? As RDK said, just do that on the two inner brackets, and you'll get the answer (after some rearranging).
Yep I got it now, thanks man
10. (Original post by RDKGames)
While this is rigorous and bound to give the correct answer, a quicker approach is to notice that the radius is 2 and centre at the origin - and so you're looking for a line which has a minimum distance from the origin as 2.

For example; the fourth equation fails this because the line crosses through and and since you have the line going through 2 axis points which are inside the circle, with minimum distance being less than 2 as a result.
Oh I see now, cheers

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