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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?

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Probably a typo, $6x$ should be $6y$. As your calculation shows, with $6x$ we do not have a circle. –  André Nicolas Oct 14 '12 at 16:45
    
That's what I thought too. –  JohnHedge Oct 14 '12 at 16:50

1 Answer 1

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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