# 2 Pre Calc problems

1. Jan 9, 2006

### Kurac

One
Use Slopes to show that A (-3,-1),B(3,3), and C(-9,8) are vertecies of a right triangle.
Two
Find an equation for the line tangent to the circle x^2+y^2=25 at the point (3,-4)

Thanks if anyone could do this that would be great.

2. Jan 9, 2006

### d_leet

For the second question the easiest way is to take the derivative, evaluate it at the point to find the slope and then write the equation of a line through that point with the slope you found.

3. Jan 9, 2006

### Kurac

could you do that for me thanks?

4. Jan 9, 2006

### Hurkyl

Staff Emeritus
Yes. But we won't.

5. Jan 9, 2006

### Little_Rascal

These can be solved very easily using basic analytic geometry. No derivation or other nasty calculus required.

For the first one:

The slope can be found using for example:
k= (y2-y1)/(x2-x1)
where:
(x1,y1) is the left endpoints coordinates
(x2,y2) is the right endpoints coordinates

if k > 0 you have a rising line
if k < 0 you have declining line

When two lines are at right angels to each other:
k1 * k2 = -1
That is, the product of the slopes equals -1.

For the second:
The point (3,-4) is located on the circle. (Do you know why?)
Try to find the slope from the circles centre to (3,-4) and then use the fact that the tangent line has to be at a right angle to the slope (Why?) to calculate the slope of the tangent line.

If you know the slope and a point you should be able to calculate the equation for the line (You probably have a formula for it).

Last edited by a moderator: Jan 9, 2006
6. Jan 9, 2006

### gaganpreetsingh

k= (y1-y2)/(x2-x1)is wrong
Slope is m = (y1-y2)/(x1-x2) not as you have given.

7. Jan 9, 2006

### Little_Rascal

Typo fixed.

8. Jan 10, 2006

### HallsofIvy

Staff Emeritus
Since this is a pre-calc problem, for number 2 try this: any line through (3,-4) can be written y= m(x-3)-4. The line tangent to x2+ y2= 25 at (3,4) must intersect it only there. For what value of m does x2+ (m(x-3)-4)2= 25 have exactly one solution for x?