Finding the line tangent to a point on a circle...

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!

Re: Finding the line tangent to a point on a circle...

Re: Finding the line tangent to a point on a circle...

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!

Re: Finding the line tangent to a point on a circle...

Quote:

Originally Posted by

**imanironmaiden** 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.

I my view this a busy work exercise. But here it is.

The divide off.

Re: Finding the line tangent to a point on a circle...

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 .

Re: Finding the line tangent to a point on a circle...

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: