Hi guys!
I was out sick for the last week or so and am catching up on all the calculus homework that I missed. There's one in particular that I'm having trouble with. The problem is to find the slope-intercept form of the line tangent to the point (6,8) on the circle x^2 + y^2 = 100. I know that the slope of a secant line is [f(x+h) - f(x)]/h and since x is 6 I changed the expression to this: [f(h+6) - 8]/h. I know that f(x) = sqrt(100 - x^2) so f(6+h) comes out to sqrt(64 - 12x - h^2) so [f(h+6) - 8]/h turns into [sqrt(64 - 12x - h^2) -8]/h and I'm not sure where to go from there since I can't seem to get h out of the denominator so I can set h to 0 without getting an undefined answer. Any help would be really appreciated. Thanks in advance!
Thanks Plato! My class hasn't touched on laws of derivatives, etc. so at this point your response looks like magic to me. Using the method I tried to use, how would I arrive at that answer and if it's not possible, why not? Thanks again for your answer!
The tangent to a circle, at a given point, is always perpendicular to the radius to that point. If the given point is (and, of course, ) then the line from the center to that point, from (0, 0) to has slope . Therefore, the slope of the tangent line is .
Another pre-calculus technique is to let the tangent line be:
Now, substitute for y into the equation of the circle:
In order for the line to be tangent to the circle, we require 1 real root, hence the discriminant must be zero:
Now, given the point (6,8) is on the line, we know:
Hence, the slope-intercept form of the tangent line is: