point of tangency

[dagr8est]

A circle of radius 4 has center at (0, 0). A point exists at (8, 0). A line is constructed from the point such that the line is tangent to the circle. Find the point(s) of tangency.

Here's what I tried:

There are two points where the line could be tangent to the circle. Let the one above the x-axis be (a, b) and the one below the x-axis be (c, d). Just dealing with (a, b) for now.

From the center of the circle, draw a radius to point (a, b). The slope of that line is m=a/b. The slope of the tangent line is -b/a.

Here's what I came up with.

circle: x^2+y^2=16
radius line: y=ax/b
tangent line y=-b(x-8)/a

This is where I'm stuck. I've tried substituting a couple different ways but no luck. Any help would be appreciated.

 

[skeeter]

let (a,b) be a point on the circle

point that passes through (8,0) and (a,b) has slope m = b/(a-8)

slope of line tangent to the circle at (a,b) has slope m = -a/b

-a/b = b/(a-8)
b^2 = 8a - a^2
a^2 + b^2 = 8a
since a^2 + b^2 = 16 ...
16 = 8a
a = 2

b = (+/-)sqrt(16 - a^2) ...
b = (+/-)2sqrt(3)

so, the line that passes through (8,0) and is tangent to x^2 + y^2 = 16 is tangent at (2, 2sqrt(3)) and (2, -2sqrt(3)).