# Find out for which values of $\lambda$ the points of the line are inside the circle

We have a line (in parameter):

$x = 2\lambda$

$y = 1-\lambda$

Find out for which values of $\lambda$ the points of the line are inside the circle of $x^2+4x+y^2-6y+5=0$

What I did:

I rewrote the circle to the form $(x+2)^2 + (y-3)^2 = 8$.

Where my problems/questions are:

• First of all, I have trouble with the parameter representation of a line, how can I rewrite this to for example $y=ax+b$ or $ax+by=c$?

• And there also is a problem with how to continue. I thought if replacing the $x$ and $y$ in the circle equation by $2\lambda$ and $y = 1-\lambda$ respectively, but I get a nonsensical answer:

$(2\lambda + 2)^2 + (1-\lambda -3)^2 = 8$

$4\lambda ^2 + 4\lambda + 4 + \lambda ^2 +4\lambda + 4 = 8$

$5\lambda ^2 + 8\lambda =0$

$\lambda ^2 + 1.6\lambda = 0$

And at this juncture I just quit because of the nonsensical answer I would get if continued. What am I doing wrong? What am I doing right? How does the parameter representation of a line work and how can I rewrite it into a different form?

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You calculation contains one mistake:

it should be $8\lambda$ instead of $4\lambda,$ as explained below:

to be inside the circle, the distance for the centre must be less than the radius.

So,$(2\lambda+2)^2+(1-\lambda -3)^2$ must be $<8$

or, $4\lambda^2+8\lambda+4+\lambda^2+4\lambda+4<8$

or $5\lambda^2+12\lambda<0$

or $(\lambda-0)\{\lambda-(-\frac{12}5)\}<0$

The product of two term $<0$, so one must $<0$ and the other $>0$

If $(\lambda-0)>0$ i.e., $\lambda>0$, then $\lambda-(-\frac{12}5)<0$ or $\lambda<-\frac{12}5$ which is impossible as $\lambda>0$.

If $(\lambda-0)<0$ i.e., $\lambda<0$,

then $\lambda-(-\frac{12}5)>0$ or $\lambda>-\frac{12}5$

So, $-\frac{12}5<\lambda<0$

Alternatively, if the equation of the circle is $x^2+y^2+2gx+2fy+c=0--->(1)$

or $\{x-(-g)\}^2+\{y-(-f)\}^2=g^2+f^2-c$

If $(h,k)$ lies inside the circle, $\{h-(-g)\}^2+\{k-(-f)\}^2<g^2+f^2-c$

or $h^2+k^2+2gh+2fk+c<0--->(2)$

Here $g=3,f=-3,c=5$ and $(h,k)$ is $(2\lambda,1-\lambda)$

We can put the values of $(h,k)$ in terms of $\lambda$ in $(2)$ to reach the same destination as in the 1st method.

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And am I right that $(1-\lambda-3)^2 = (-\lambda-2)^2?$ – JohnPhteven Nov 2 '12 at 16:48
@ZafarS, yes $=(\lambda+2)^2$ too. Modified the answer. – lab bhattacharjee Nov 2 '12 at 16:50
However, when I try to eliminate $y$, so making it like this: $(x+2)^2 + (0,5x+2)^2=*$, I get the answers $x=0$ and $x=-4.8$, however that should be $x=-2,4$ too right? Can I use the method of elimination too? – JohnPhteven Nov 2 '12 at 17:10
*=8, I can't edit it after 5 minutes for some reason.. – JohnPhteven Nov 2 '12 at 17:16
@ZafarS, it's not '=', but '<' – lab bhattacharjee Nov 2 '12 at 17:29

As for converting the parametric form to a regualar form, it goes as follows. First express either $x$ or $y$ completely in terms of the parameter. Here, for example, $y=1-\lambda$, so express $\lambda=1-y$ and replace this value of $\lambda$ in the other equation. Doing so, we get $x=2(1-y)$ or $y=\frac{-x}{2}+1$. That's all.

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