Answer:
First, plot your circle. the Center is (0, 3) and the radius is [tex]\sqrt{34}[/tex]
Now to find your slope move from the (0, 3) to (5, 0)
[tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{0 - 3}{5 - 0} = -\frac{3}{5}[/tex]
The slope of a perpendicular line is opposite and reciprocal, so the slope is [tex]\frac{5}{3}[/tex]
The equation of our line tangent to the circle at the point (5, 0) is [tex]y = \frac{5}{3}x + b[/tex]
Now substitute in the point (5, 0) to solve for b
[tex]0 = \frac{5}{3}(5) + b\\0 = \frac{25}{3} + b\\b = -\frac{25}{3}[/tex]
Therefore, the equation of the line tangent to the circle is
[tex]y = \frac{5}{3}x - \frac{25}{3}[/tex]
Step-by-step explanation:
