1. ## Due tomorrow:

This is for a calculus class (but we're doing the preliminary chapter, and I can't actually use any calculus).

1. Draw the unit circle and plot the point P = (4,2). Find the equation of a line tangent to the unit circle passing through the point, P.

2. A rancher has 300 ft. of fence to enclose two adjacent pastures.
a) Write the total are A of the two pastures as a function of x.

For this, I used the equations for the area and perimeter based on the diagram of the pasture layout we were given, A = x2y, and 300 = 3x = 4y, and got the function A(x) = 150x - 1.5x^2:

300 = 3x + 4y
300 - 3x = 4y
75 - .75x = y

A = x2(75 - .75x)
A = x(150 - 1.5x)
A(x) = 150x -1.5x^2

Firstly, did I do this right?

b) Find the dimensions that yield the maximum amount of area for the pastures by completing the square.

2. A geometric solution to the first problem: Draw a line from (4,2) to (0,0). The length of this line is $\sqrt{20}$. Now, the two tangent lines will intersect the unit circle at a point 1 unit from the origin, and because the tangent lines will be perpendicular to the radius drawn to the points, the distance from (4,2) to either of these points will be $\sqrt{(\sqrt{20})^2 - 1^2} = \sqrt{19}$. To find the points, then, we can use the intersection of the two circles $(x-4)^2 + (y-2)^2 = 19$ and $x^2 + y^2 = 1$.

3. So, if I did my math right (which may or may not be the case) (0,1) is a point of tangency, and y = 1/4x + 1 is an equation which answers the question.

4. Originally Posted by justaguy
So, if I did my math right (which may or may not be the case) (0,1) is a point of tangency, and y = 1/4x + 1 is an equation which answers the question.
I don't think that is right. Both points of tangency lie on the line y = 1/2 - 2x. You can use this fact in conjunction with the unit circle to produce a quadratic in x or y, which you can then use to find the actual points of tangency.

5. Yeah, that can't be right, actually. The line from (0,0) to (0,1) isn't perpendicular to the line from (0,1) to (4,2). How do you find y = 1/2 - 2x?

On the plus side, I've figured out the other problem.

6. This is my work for the intersection of the circles:

$(x-4)^2 + (y-2)^2 = 19$

$x^2 - 8x + 16 + y^2 - 4y + 4 = 19$

$(x^2 + y^2) - 8x - 4y + 20 = 19$

$1 - 8x - 4y + 20 = 19$

$21 - 8x - 4y = 19$

$21 = 19 + 8x + 4y$

$8x + 4y = 2$

$4y = 2 - 8x$

$y = \frac{1}{2} - 2x$

Now back to the unit circle:

$x^2 + y^2 = 1$

$x^2 + \left(\frac{1}{2} - 2x \right)^2 = 1$

$x^2 + \frac{1}{4} - 2x + 4x^2 = 1$

$5x^2 - 2x + \frac{1}{4} = 1$

$5x^2 - 2x - \frac{3}{4} = 0$

$20x^2 - 8x - 3 = 0$

To find the two points of tangency, substitute these x-values back into the line $y = \frac{1}{2} - 2x$.

7. Hey, I really appreciate the help.

I'd plugged the equation you gave me, $y = 1/2 - 2x$, into the formula for the unit circle, and I'm with you up until . How did you get that from ?

8. I just multiplied both sides of the equation by 4. You don't actually need to do that in order to solve the equation, but I wanted to make it look cleaner.

9. Of course, how silly of me. I think I've finally got it. Thanks a lot!