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

Printable View

September 11th 2012, 01:10 PM
imanironmaiden

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!
September 11th 2012, 01:37 PM
Plato

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

Quote:

Originally Posted by imanironmaiden

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.

The derivative is $2x+2yy^{\prime}=0$ so $y^{\prime}=\frac{-x}{y}$ .
So use $(6,8)$ to find the slope $y^{\prime}$ .
September 11th 2012, 02:26 PM
imanironmaiden

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!
September 11th 2012, 02:36 PM
Plato

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.

$\frac{{\sqrt {100 - \left( {{x^2} + 2hx + {h^2}} \right)} - \sqrt {100 - {x^2}} }}{h} = \frac{{ - 2hx - {h^2}}}{{h\left( {\sqrt {100 - \left( {{x^2} + 2hx + {h^2}} \right)} + \sqrt {100 - {x^2}} } \right)}}$

The $h's$ divide off.
September 11th 2012, 02:57 PM
HallsofIvy

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 $(x_0, y_0)$ (and, of course, $x_0^2+ y_0^2= 100$ ) then the line from the center to that point, from (0, 0) to $(x_0, y_0)$ has slope $\frac{y_0}{x_0}$ . Therefore, the slope of the tangent line is $-\frac{x_0}{y_0}$ .
September 11th 2012, 03:26 PM
MarkFL

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

Another pre-calculus technique is to let the tangent line be:

$y=mx+b$

Now, substitute for y into the equation of the circle:

$x^2+(mx+b)^2=100$

$x^2+m^2x^2+2bmx+b^2=100$

$\left(m^2+1 \right)x^2+(2bm)x+\left(b^2-100 \right)=0$

In order for the line to be tangent to the circle, we require 1 real root, hence the discriminant must be zero:

$(2bm)^2-4\left(m^2+1 \right)\left(b^2-100 \right)=0$

$4b^2m^2-4\left(b^2m^2-100m^2+b^2-100 \right)=0$

$400m^2-4b^2+400=0$

$100m^2-b^2+100=0$

Now, given the point (6,8) is on the line, we know:

$b=8-6m$

$100m^2-(8-6m)^2+100=0$

$100m^2-64+96m-36m^2+100=0$

$64m^2+96m+36=0$

$16m^2+24m+9=0$

$(4m+3)^2=0$

$m=-\frac{3}{4}$

$b=\frac{25}{2}$

Hence, the slope-intercept form of the tangent line is:

$y=-\frac{3}{4}x+\frac{25}{2}$