Question

How do you find the equation of the tangent line to the curve `x^2 + y^2= 169` at the point (5,-12)?

Answer 1

Find the slope of the line, `m=5/12`, then use the point `(5,-12)` to get the equation `y=5/12x-169/12`.

Explanation:

Here are three ways to find the slope of the tangent line:

Geometry and algebra (no calculus)

The graph of `x^2+y^2=169` is a circle centered at `(0,0)` (with radius `13`).

The line tangent to the circle at `(5,-12)` is perpendicular to the radius drawn between the origin and the point `(5,-12)`. The slope of that radius is `-12/5`

So the slope of the tangent line is `5/12` (perpendicular lines that have slopes have slopes that are opposite reciprocals.)

Calculus explicit solution

Solve `x^2+y^2=169` for `y` to get

`y=+-sqrt(169-x^2)`

We are interested in the part of the graph that has negative `y` values, so we want

`y=-sqrt(169-x^2)`.

Differentiate to get `y'=x/sqrt(169-x^2)`

At `x=5` we get a slope of `m = 5/sqrt(169-5^2) = 5/sqrt144 = 5/12`

Calculus implicit solution

`x^2+y^2=169`. So,

`2x+2y dy/dx =0` and

`dy/dx = -x/y`.

At `(5,-12)`, we get `m=-(5)/(-12) = 5/12`