Let $(x_0,y_0)$ be the point on the circle $x^2+y^2=1$ where the line $L$ from $(1,3)$ is tangent to this circle. The slope of the line $L$ is \[m=\frac{y_0-3}{x_0-1}.\] On the other hand, the line $L'$ passing from the origin and $(x_0,y_0)$ is perpendicular to $L$. The slope of $L'$ is $m'=\frac{y_0}{x_0}$. Thus \[-1=m \times m'= \frac{y_0-3}{x_0-1} \times \frac{y_0}{x_0}=\frac{y_0^2-3y_0}{x_0^2-x_0}=\frac{1-x_0^2 - \pm 3 \sqrt{1-x_0^2}}{x_0^2-x_0},\] because $y_0^2=1-x_0^2.$ So \[x_0-x_0^2=1-x_0^2- \pm 3 \sqrt{1-x_0^2}\] \[\Rightarrow 1-x_0=\pm 3 \sqrt{1-x_0^2} \Rightarrow (1-x_0)^2=9(1-x_0^2)\] \[(1-x_0) (9(1+x_0)-(1-x_0))=0 \Rightarrow (1-x_0) (8+10x_0)=0.\] Hence we get (i) $x_0=1$ or (ii) $x_0=-\frac{8}{10}=-\frac{4}{5}$.
For $x_0=1$ we have $y_0=0$, and the line passing through $(1,0)$ and $(1,3)$ is the vertical line $x=1$.
For $x_0=-\frac{4}{5}$, we get $y_0=\pm \sqrt{1-(\frac{4}{5})^2}=\pm \frac{3}{5}$. One can easily check that $(-\frac{4}{5}, -\frac{3}{5})$ does not satisfy \[-1=m \times m'= \frac{y_0-3}{x_0-1} \times \frac{y_0}{x_0},\] but the point $(-\frac{4}{5}, \frac{3}{5})$ does satisfy this condition. The equation of the line tangent to the circle at the poin $(-\frac{4}{5}, \frac{3}{5})$ passing from $(1,3)$ is
\[y-3=\frac{3-\frac{3}{5}}{1-(-\frac{4}{5})}(x-1) \Rightarrow y=3+\frac{4}{3}(x-1).\]
So there are two different lines tangent to the given circle from $(1,3)$:
(i) $x=1$ and (ii) $y=3+\frac{4}{3}(x-1)$.
Theresa Lee
