Help in maths p1!

[immortal]

Find the values of k for which the line y=kx-2  meets da curve y²=4x-x²

[past-papers]

simultaneous equations:

[nid404]

Find the values of k for which the line y=kx-2  meets da curve y²=4x-x²

Post the paper up

[past-papers]

wait, actually you're missing something --> post the whole question

[immortal]

I posted da whole question which i got 4 ma Mocks.da only thing i missed out waz dat it waz a [4] marks question.

[student]

I didnt do p1, but..

y²=4x-x² is a circle, isnt it?
With 0=<x=<4.

And y=kx-2 would pass through this circle at k>0 to the infinity.

[tmisterr]

simultaneous equations:

You probably are missing something but if not continuing from past papers equation

k²x²-x²-4kx-4x+4=0 so as you know if it meets the curve it must either have a repeated root (b²=4ac or it b²>4ac

so (-4k-4)²>or=4*(k²+1)*4
solving this equation gives you k>or=0.

[nid404]

yup that what i did and concluded there is something missing

[immortal]

y=kx-2  meets da curve y²=4x-x², so it has a root.

(kx-2)²=4x-x²
k²x²-4kx+4=4x-x²
k²x²+x²-4kx-4x+4=0
(k²+1)x²-(4k+4)x+4=0

as da eqn haz 1 root b²-4ac=0
(4k+4)²-(4*(k²+1)*4)=0
16k²+32k+16-(16k²+16)=0

but i get 32k=0 which is probably wrong!
any ideas Guyz

[student]

Find the values of k for which the line y=kx-2  meets da curve y²=4x-x²

There's not one root.
And y=kx-2 does meet y²=4x-x² at any values of k>0 or equal to 0..
Otherwise there might be some special 'p1 method' for it?