# Problem with finding the equations of the lines tangent to a certain circle

This is a long question, and might seem like a repost of my earlier questions, but it isn't, hear me out:

In my book is written:

The equation of the line tangent to the circle $x^2+y^2=r^2$ in the point $P(a,b)$ is $ax+by=r^2$. This is only if $P$ is a point on the circle. However, if $P$ is not a point on the circle, you can find the line which goes through the points of tangency. Here an example of such a line (called poollijn in Dutch, don't know the English name, I'll keeping reffering to it as poollijn ):

Now we have the following question:

Find the equations of the lines from $P(0,6)$ tangent to the circle $x^2+y^2=4x+4$

I have posted a question about this one on the forum before, and I solved it afterwards by simply eliminating $y$. However, I ALSO wanted to know how to solve this problem using the poollijn. Sadly, I haven't yet. When I went to check the correction sheet, I saw that I made a mistake right at the beginning, these are the first steps:

$xa+by=4x+4$

$x.0+y.6 = 2x + 2.0 +4$

$6y = 2x+4$

$y= \dfrac{1}{3}x + \dfrac{2}{3}$ (poollijn)

What I don't understand, is the second step. Why does the RHS suddenly change from $4x+4$ to $2x+2.0+4$? It must be right too, since the final answers you get by using this method (first finding the poollijn, then the intersection with the circle, and then you have 2 points per line, so you can make the equations) are correct. So what am I seeing incorrectly. Why the change from $4x+4$ to $2x+4$

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The square of the radius of the circle is not $4x+4$. You need to rewrite the equation of the circle as $(x-2)^2+y^2=8$, which shows it is centered at $(2,0)$ with radius $\sqrt 8$. So let $x'=x-2$. You want the equation from $P=-(2,6)$ in the $(x',y)$ system to the circle.

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I thought that too, but how can one get the correct poollijn (what is the English name for it, can find it on the internet) any other way? – JohnPhteven Oct 23 '12 at 15:27
According to Google Translate, poollijn is Dutch for polar line. Is that the correct original language? – Alan Guo Oct 23 '12 at 15:35
Yes indeed. So it's polar line? That's sounds logical.. – JohnPhteven Oct 23 '12 at 15:36
@ZafarS: I don't see how you are finding the poollijn from the coordinates of $P$, nor do I see how the answer sheet is doing its work at all. – Ross Millikan Oct 23 '12 at 15:45

(1)From article#$151,372$ of this or article#$85,86$ of this, the equation of the tangent at $P(x_1,y_1)$ to the curve $S(x,y)=ax^2+2hxy+by^2+2gx+2fy+c=0$ when P lies on the curve is

$T(x,y)=ax_1x+h(y_1x+x_1y)+by_1y+g(x+x_1)+f(y+y_1)+c=0$

Here, $a=b=1,h=0,g=-2,f=0,c=-4$

This is exactly why $4x$ becomes $2(x+0)$.

(2)Form article#$389$ of this or article $92$ of this , if $P(x_1,y_1)$ does not lie on the curve, we can draw a pair of tangents from P to the curve $S(x,y)=0$, given by $S(x,y)\cdot S(x_1,y_1)=T^2(x,y)$

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