SOLUTION: Find the slope of the line tangent to a circle with center o at (5,3) at point B (-2,5). I got: -7/2 Is this right? Thanks

Find the slope of the line tangent to a circle with center 
O at (5,3) at point B (-2,5). 
I got: -7/2 
Is this right? 
Thanks

I think you missed a sign. Let's draw the picture:



We can tell that at least you have the sign wrong because lines
which go UPHILL TO THE RIGHT have slopes which are POSITIVE and
lines which go DOWNHILL TO THE RIGHT have slopes which are NEGATIVE.
The tangent line at B goes UPHILL TO THE RIGHT, so it must have a
POSITIVE slope, so your negative slope has to be wrong.

First let's find the slope of radius OB:

Use the slope formula:

     y2 - y1
m = —————————
     x2 - x1

where O(x1, y1) = O(5, 3) and B(x2, y2) = B(-2, 5)

     (5) - (3)      2       2
m = ——————————— = ———— = - ——— 
    (-2) - (5)     -7       7 

The radius goes DOWNHILL TO THE RIGHT, so we expected
that to have a NEGATIVE slope, and -2/7 is indeed 
negative.

A tangent line is perpendicular to the radius drawn to
the point of tangency.  So the tangent line is 
perpendicular to a line with slope -2/7.  The rule
for finding the slope of a line perpendicular to another
line is:

1. Invert the slope.
2. Change its sign.

So we take the slope of the radius OB, which is -2/7,
then 

1. we invert it and get -7/2

and then,

2. we change the sign of -7/2 to +7/2

So the correct answer is +7/2.

Always use the fact that lines with positive slopes
go UPHILL to the RIGHT and lines with negative slopes
go DOWNHILL to the RIGHT.  That way you won't miss
any more signs for slopes if you draw the graphs.
You might miss the number but you won't miss the sign!

Edwin