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?

Answer 1

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

Answer 2

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

Answer 3

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.