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

1 Answer
Apr 21, 2016

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#