# Find point of contact when tangent and circle equations are given.

If $L=lx+my+n=0$ is tangent to circle $x^2+y^2+2gx+2fy+c=0$ then find point of contact.

I know that equation of tangent is $S_{1}=xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$ anf rearranging I obtain $S_1=(g+x_1)x+(f+y_1)y + gx_1+fy_1+c=0$.

By comparing coefficients it should give the point of contact: $x_1=l-g$, $y_1=m-f$. But when I use this method in exercise it gives me wrong a point of contact.

Where I am going wrong? Help.

The linear equations $$L=0$$ and $$S_1=0$$ are equivalent iff the coefficients are proportional. So, you should use the equations $$\frac{g+x_1}{l}=\frac{f+y_1}{m}=\frac{gx_1+fy_1+c}{n},$$ and by solving the linear system $$\begin{cases} \displaystyle\frac{g+x_1}{l}=\frac{f+y_1}{m}\\ \displaystyle\frac{gx_1+fy_1+c}{n}=\frac{f+y_1}{m} \end{cases}$$ that is $$\begin{cases} mx_1-ly_1=lf-mg\\ mgx_1+(mf-n)y_1=nf-mc \end{cases}$$ we find $$x_1 = \frac{lf^2-mgf+ng-lc}{lg+mf-n}\quad,\quad y_1 = \frac{mg^2-lfg+nf-mc}{lg+mf-n}.$$
• The linear equations $L=0$ and $S_1=0$ are equivalent if the coefficients are proportional. Commented May 23, 2017 at 7:13