Coordinates Of Circle Center Further Maths GCSE

Absolutely no idea how to do this one...

Absolutely no idea how to do this one...

Absolutely no idea how to do this one...

$3(x^2+y^2-4x-3y)+1=0$

Now complete the squares on x and y inside the bracket.

Also this one, is it just trial and error?

Thanks a lot! I got that far before getting myself confused

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

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).

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 $(0,\sqrt{2})$ and $(\sqrt{2},0)$ and since $\sqrt{2}<2$ you have the line going through 2 axis points which are inside the circle, with minimum distance being less than 2 as a result.