Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $2x+4$?

share|improve this question
1  
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 $2x+2.0+4$ –  ZafarS 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. –  ZafarS Oct 23 '12 at 12:00

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer
    
@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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.