# Urgent - Find the equation of the lines tangent to a circle

Question: 'Find the equation of the lines from point $P(0,6)$ tangent to the circle $x^2+y^2=4x+4$.

So what I did firstly is rewrite it to the form $(x-2)^2 + y^2 = 8$, and I saw that point $P$ is not on the circle. I learned that the equation of the line tangent to the circle $x^2+y^2=r^2$ from the point $P(a,b)$ is $xa+yb=r^2$

$xa+yb=4x+4$

$x.o+y.6=2x+2.0 +4$ (This is the step I don't understand)

$6y=2x+4$

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

So basically, my question is: Why did the $4x+4$ change into the $2x+4$?

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Use that the tangent of a circle is perpendicular to the radius of the same tangent point. –  Berci Oct 23 '12 at 10:38
Note that your circle is not of the form $x^2+y^2=r^2$. But why the urgency? Are you currently taking an exam? If this is homework, would you be kind enough to edit in the "homework" tag? –  Gerry Myerson Oct 23 '12 at 11:42
It is not homework, but I'm at a crucial point of my self-study and this question has been bothering me for weeks now, simply because I don't understand the step that I noted (why did $4x+4$ change into $2x+2.0+4$ –  JohnPhteven Oct 23 '12 at 11:46
As I said, your circle is not of the form $x^2+y^2=r^2$. Thus, your identification of $r^2$ with $4x+4$ was unjustified. So $4x+4$ didn't change into anything --- $4x+4$ was never there in the first place. You tried to apply a formula in a setting in which is was not applicable. Read the correct solutions people are posting to see how it should be done. –  Gerry Myerson Oct 23 '12 at 11:59
Oh, thank you. Well, again the correction model is wrong, unbelievable. –  JohnPhteven Oct 23 '12 at 12:00

The equation of any line passing through $(0,6)$ can be written as $\frac{y-6}{x-0}=m$ where $m$ is the gradient, So, $y=mx+6$

If the $(h,k)$ be the point of contact, then $k=mh+6$ and $h^2+k^2-4h-4=0$

Replacing $k$ in the 2nd equation, $h^2+(mh+6)^2-4h-4=0$

or $(1+m^2)h^2+2h(6m-2)+32=0$, it is a quadratic equation in $h,$

For tangency, both the root should be same, so, $4(6m-2)^2=4\cdot(1+m^2)32$

$(6m-2)^2=(1+m^2)32, (3m-1)^2=8(1+m^2) \implies m^2-6m-7=0$

So, $m=7,-1$

If $m=7,\frac{y-6}{x-0}=7, 7x-y+6=0$

If $m=-1,\frac{y-6}{x-0}=-1, x+y-6=0$

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@ZafarS, the equation of the tangent, you've learnt is applicable for points$(a,b)$ lying on the circle, but here $(0,6)$ does not.In fact, we shall always find a pair of tangents in this case. –  lab bhattacharjee Oct 23 '12 at 12:46
@Zafars, could you please look into : math.stackexchange.com/questions/219453/… for $4x+4 -> 2x+4$ –  lab bhattacharjee Oct 24 '12 at 4:28

Your circle has center $c = (2,0)$ and radius $r = 2\sqrt{2}$.

If $(x,y)$ is on the circle, the line $$(x,y) + (y - c_y,c_x - x)t$$ is a tagent, since $(y - c_y, c_x - x)\cdot(x - c_x, y - c_y) = 0$. Now, since you additionally want $p = (p_x,p_y)$ to be on that line, you get the equation $$(x,y) + (y - c_y,c_x - x)t = (p_x,p_y)$$

Alternatively, you can directly characterize the lines by requiring that $(x,y) - (p_x,p_y)$ is normal to the radius through $(x,y)$, i.e. to $(x-c_x,y-c_y)$. That way, you get $$(x-p_x,y-p_y)\cdot(x-x_c,y-c_y) = 0$$

Note that these equation don't really describe the line, they describe a second point $(x,y)$, beside $P$, on the line. But once you have that, the line is easily described. Also note that these equation themselves don't force $(x,y)$ to lie on the circle. They're always meant to be used together with your original equation for the circle.

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I appreciate your answer, but I edited my question to make my problem more clear. Would you mind taking a look? –  JohnPhteven Oct 23 '12 at 11:03