<s>[INST] Analytic Geometry: Circle Given is a line with parametric equation:
$ x = 2 \lambda $
$ y = 1-\lambda $
Find out for which values of $\lambda$ the line is inside the circle of $x^2+4x+y^2-6x+5=0$
My attempt at solving this:
$x^2+4x+y^2-6x+5=0$ 
$x^2-2x+y^2+5=0$
$ (x-1)^2 -1 + y^2+5=0$
$ (x-1)^2  + y^2  =-4$
And that's where I'm stuck, this isn't correct. Can anyone point my in the right direction?
 [/INST] I will assume that we have a typo, and $x^2+4x+y^2-6x+5=0$ is intended to be $x^2+4x+y^2-6y+5=0$. Two reasons for this: (i) it would be strange to have two "$x$" terms and (ii) the given equation is not satisfied by any (real) pair $(x,y)$.
The modified equation can be rewritten as $(x+2)^2+(y-3)^2=8$, a circle with centre $(-2,3)$ and radius $\sqrt{8}$.
So we want the distance from $(2\lambda,1-\lambda)$ to $(-2,3)$ to be $\lt\sqrt{8}$. 
Equivalently, we want to solve the inequality
$$(2\lambda+2)^2+(1-\lambda-3)^2\lt 8.$$
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