*Find the equation of the tangent to the circle above that passes through the beginning of axis O (0,0)
1. Center and radius of the circle:
\(\begin{array}{|rcll|} \hline x^2+y^2-6x-2y+8 &=& 0 \\ x^2-6x+y^2-2y &=& -8 \\ (x-\frac{6}{2})^2-3^2 +(y-\frac{2}{2})^2 -1 &=& -8 \\ (x-3)^2 +(y-1)^2 &=& 9+1-8 \\ (x-3)^2 +(y-1)^2 &=& 2 \\\\ Center = C(3,1) && Radius\ r= \sqrt{2} \\ \hline \end{array}\)
\(OC = \sqrt{x_c^2+y_c^2} =\sqrt{3^2+1^2}= \sqrt{10}\)
\(\begin{array}{|rcll|} \hline \tan(\alpha) &=& \frac{y_c}{x_c} \\ &=& \frac{1}{3} \\ \tan(\beta) &=& \frac{r}{\sqrt{OC^2-r^2}} \\ &=& \frac{\sqrt{2}}{\sqrt{10-2}} \\ &=& \frac{1}{2} \\ \hline \end{array} \)
2. Tangent above slope m:
\(\begin{array}{|rcll|} \hline m=\tan{(\alpha+\beta)} &=& \frac{\tan(\alpha)+\tan(\beta)} {1-\tan(\alpha)\tan(\beta)} \\ m &=& \frac{\frac{1}{3}+\frac{1}{2}} {1-\frac{1}{3}\cdot \frac{1}{2}} \\ m &=& 1 \\\\ y &=& mx +b \quad & | \quad m = 1 \qquad b = 0 \\ \mathbf{y} &\mathbf{=}& \mathbf{x} \\ \hline \end{array} \)
3. Tangent below slope m:
\(\begin{array}{|rcll|} \hline m=\tan{(\alpha-\beta)} &=& \frac{\tan(\alpha)+\tan(\beta)} {1+\tan(\alpha)\tan(\beta)} \\ m &=& \frac{\frac{1}{3}-\frac{1}{2}} {1+\frac{1}{3}\cdot \frac{1}{2}} \\ m &=& -\frac{1}{7} \\\\ y &=& mx +b \quad & | \quad m = -\frac{1}{7} \qquad b = 0 \\ \mathbf{y} &\mathbf{=}& \mathbf{-\frac{1}{7}\cdot x} \\ \hline \end{array}\)
