I would have started with partial differentiation, I think the y-intercept bit is there to get rid of one of the solutions at the end.
Ok this turned out way longer than I thought it was going to, so it's may not be the best way, but it works, so I hope it still helps in someway (or someone else may be able to see a better way).
Anyway, using implicit differentiation
^{2}+2(y+1)^{2}\right) & =\frac{d}{dx}(4)<br />\\ 2(x-2)+4(y+1)\frac{dy}{dx} & =0<br />\\ \frac{dy}{dx} & =\frac{-2(x-2)}{4(y+1)}<br />\\ \frac{dy}{dx} & =\frac{2-x}{2(y+1)}<br />\end{alignedat})
Now we know that at
)
this will be parallel to
, that is share the same gradient, so that is
at
} & =1<br />\\ 2-a & =2b+2<br />\\ a & =-2b<br />\end{alignedat})
EDIT: for those who need this question, the better method from here onwards is to sub (a,b) and a=-2b into the tangent equation, as pointed out by Bozo
Substituting this back into the equation for the circle, at
and
^{2}+2(b+1)^{2} & =4<br />\\ (-2b-2)^{2}+2(b^{2}+2b+1) & =4<br />\\ (-(2b+2))^{2}+2b^{2}+4b+2 & =4<br />\\ 4b^{2}+8b+4+2b^{2}+4b+2 & =4<br />\\ 6b^{2}+12b+2 & =0<br />\end{alignedat})
(2)}}{2(6)}<br />\\ & =\frac{-12\pm\sqrt{96}}{12}<br />\\ & =\frac{-12\pm\sqrt{16\times6}}{12}<br />\\ & =\frac{4(-3\pm\sqrt{6})}{12}<br />\\ & =\frac{-3\pm\sqrt{6}}{3}<br />\end{alignedat})
, so for
For
BUT these points must satisfy our extra condition for our y-intercept for the tangent.
So using you equation of the tangent, lets sub in a and check if we get b.
For
)
 & =2+\frac{2\sqrt{6}}{3}+3\sqrt{\frac{2}{3}}-3<br />\\ & =-1+\frac{2\sqrt{6}}{3}+\frac{3\sqrt{6}}{3}<br />\\ & =-1+\frac{5\sqrt{6}}{3}<br />\end{alignedat})
Now this is
not the value of b for this solution, so it is not the answer.
For
)
 & =2-\frac{2\sqrt{6}}{3}+3\sqrt{\frac{2}{3}}-3<br />\\ & =-1-\frac{2\sqrt{6}}{3}+\frac{3\sqrt{6}}{3}<br />\\ & =-1+\frac{\sqrt{6}}{3}<br />\end{alignedat})
Now this
is the value of b for this solution, so this is your answer.
I feel there should be a quicker way to do this, I haven't picked up on it yet if there is, but someone else may be able to.
Anyway hope that helps
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