Consider the circle of radius 5 centered at (0,0), how do you find an equation of the line tangent to the circle at the point (3,4)?

1 Answer
Apr 20, 2015

Using calculus:

the equation of the circle is #x^2+y^2 = 25#

At the point #(3,4)#, we are on the upper semicircle, whose equation is: #y = sqrt(25-x^2)#

So #dy/dx = 1/(2sqrt(25-x^2)) (-2x) = (-x)/sqrt(25-x^2)#

At #(3,4)#, we get #dy/dx = (-3)/4#

So the slope of the line is #-3/4# and the line contains the point #(3,4)#.

#y=-3/4 x +25/4#

Without calculus:

The tangent is perpendicular to the radius connecting #(0,0)# and #(3,4)#.

The slope of the radius is #4/3#, so the slope of the tangent is #-3/4#