Find the condition that the line $lx+my+n=0$ is a tangent to the circle $x^2+y^2=r^2$

Question

Find the condition that the line $lx+my+n=0$ is a tangent to the circle $x^2+y^2=r^2$.

My Attempt: \begin{align}lx+my+n&=0\\ y&=\dfrac {-n-lx}{m}\end{align}

Now, \begin{align}x^2+y^2&=r^2\\ x^2+(\dfrac {-n-lx}{m})^2&=r^2\\ x^2+\dfrac {n^2+2nlx+l^2x^2}{m^2} &=r^2\\ x^2(m^2+l^2) + 2nlx +n^2-m^2r^2&=0\end{align}

How do I continue further?

You may want to check:https://math.stackexchange.com/questions/2254073/the-line-y-mxc-is-a-tangent-to-x2y2-a2-if — Peter Szilas, May 10 '17 at 14:42

score 3 · Accepted Answer · answered May 10 '17 at 14:16

3

The line is tangent to the circle, if its distance to the center $(0,0) $ equals the radius $r :$

$$\frac{|l.0+m.0+n|}{\sqrt {l^2+m^2}}=r $$

or $$n^2=r^2 (l^2+m^2) $$

answered May 10 '17 at 14:16

hamam_Abdallah

62,951

score 1 · Answer 2 · answered May 10 '17 at 14:13

1

Compute the discriminant of your last expression and make it equal to zero. That's the condition for tangency.

answered May 10 '17 at 14:13

Could you please show me the calculations? I am.not getting that. – pi-π May 10 '17 at 14:16
@AlbertEinstein there's not a lot to show. The discriminant equal to zero implies that there's only one value of $x$ that satisfies the equation of the line and the equation of the circle. That's the key. – May 10 '17 at 14:20
BTW @Salahamam_Fatima 's answer is more elegant and intuitive. I was just working over your approach – May 10 '17 at 14:22

score 0 · Answer 3 · answered May 10 '17 at 15:13

Equation of tangent in point of contact $(x_1,y_1)$ form can be written as $$xx_1+yy_1-r^2=0$$ Since $lx+my+n =0 $ also represent the same tangent . $$\implies \frac{x_1}{l}=\frac{y_1}{m}=\frac{-r^2}{n}$$ Now you can find point of contact ,and it lies on circle so it satisfies the equation of circle .That will be required condition.

Find the condition that the line $lx+my+n=0$ is a tangent to the circle $x^2+y^2=r^2$

3 Answers3