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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.

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1 Answer 1

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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 $$x_1 = \frac{lf^2-mgf+ng-lc}{lg+mf-n}\quad,\quad y_1 = \frac{mg^2-lfg+nf-mc}{lg+mf-n}.$$

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  • $\begingroup$ Looks correct. But I wonder why that equation is used. $\endgroup$
    – Fawad
    Commented May 23, 2017 at 7:08
  • $\begingroup$ The linear equations $L=0$ and $S_1=0$ are equivalent if the coefficients are proportional. $\endgroup$
    – Robert Z
    Commented May 23, 2017 at 7:13
  • $\begingroup$ @Fawad Any further doubts? $\endgroup$
    – Robert Z
    Commented May 23, 2017 at 7:23
  • $\begingroup$ Nope. I understood. Another way to solve is that point of contact is foot of perpendicular to centre of circle on tangent line. In equation of foot of perpendicular all variables are known except point of contact. Both gives same equation. Thanks! $\endgroup$
    – Fawad
    Commented May 23, 2017 at 7:28
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    $\begingroup$ @F.N. I just solved the linear system. See my edit. $\endgroup$
    – Robert Z
    Commented Jan 1, 2021 at 9:50

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