Question

Find an equation of a line tagent to the circle x^2+(y-3)^2=34 at the point (5,0)

Answer 1

Answer:

First, plot your circle. the Center is (0, 3) and the radius is $\sqrt{34}$

Now to find your slope move from the (0, 3) to (5, 0)

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{0 - 3}{5 - 0} = -\frac{3}{5}$

The slope of a perpendicular line is opposite and reciprocal, so the slope is $\frac{5}{3}$

The equation of our line tangent to the circle at the point (5, 0) is $y = \frac{5}{3}x + b$

Now substitute in the point (5, 0) to solve for b

$0 = \frac{5}{3}(5) + b\\0 = \frac{25}{3} + b\\b = -\frac{25}{3}$

Therefore, the equation of the line tangent to the circle is

$y = \frac{5}{3}x - \frac{25}{3}$

Step-by-step explanation: