Question 44641
{{{x^2+4.y^2-4.x-8.y+4=0}}}
Rearrange to give a quadratic equation in terms of y:
{{{4y^2-8y+(x^2-4x+4)=0}}}
Now solve the quadratic using {{{y = (-b +- sqrt( b^2-4ac ))/(2a) }}}, 
where
{{{a=4}}}
{{{b=-8}}}
{{{c=x^2-4x+4}}}
This should give
{{{y = (8 +- sqrt( 64-16(x^2-4x+4)))/8 }}}
{{{x^2-4x+4}}} has two equal roots {{{(x-2)^2=(x^2-4x+4)}}}, so the quadratic solver for y becomes:
{{{y = (8 +- sqrt( 64-16(x-2)^2))/8 }}}
{{{y = 1 +- sqrt( 64-16(x-2)^2)/8 }}}
There are two existing solutions for y:
{{{y = 1 + sqrt( 64-16(x-2)^2)/8 }}} and {{{y = 1 - sqrt( 64-16(x-2)^2)/8 }}}.
I have plotted both of the lines below for you.

{{{ graph( 300, 200, 0, 4, -10, 10, 1 + sqrt( 64-16(x-2)^2)/8, 1 - sqrt( 64-16(x-2)^2)/8) }}}


I hope this helps.
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk
Adam