Solve 2 Pre Calc Probs: Slopes & Tangent to Circle

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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.
 
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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.
 
could you do that for me thanks?
 
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).
 
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k= (y1-y2)/(x2-x1)is wrong
Slope is m = (y1-y2)/(x1-x2) not as you have given.
 
gaganpreetsingh said:
k= (y1-y2)/(x2-x1)is wrong
Slope is m = (y1-y2)/(x1-x2) not as you have given.

Typo fixed.
 
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?