Question

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

Answer 1

First, plot your circle. the Center is (0, 3) and the radius is
`√(34)`

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

`(y_(2)-y_(1))/(x_(2)-x_(1)) = (0 - 3)/(5 - 0) = -(3)/(5)`

The slope of a perpendicular line is opposite and reciprocal, so the slope is
`(5)/(3)`

The equation of our line tangent to the circle at the point (5, 0) is
`y = (5)/(3)x + b`

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

`0 = (5)/(3)(5) + b\\0 = (25)/(3) + b\\b = -(25)/(3)`

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

`y = (5)/(3)x - (25)/(3)`

Explanation: