Question

Does anyone know how to solve this.... write the slope intercept of the line tangent to (x-2)^2+(y+3)^2=9 at the point (-1,-3).

Answer 1

A nice trick, write it like this:

\[\Large (x_0-2)(x-2)+(y_0+3)(y+3)=9 \]
and substitute for \( x_0, y_0\) your respective values of the point you want to make tangent.

Answer 2

This safes you of doing the implicit differentiation, you might later encounter the same method again when it comes to finding the polar of a circle.

Answer 3

slope is undefined at that point

Answer 4

true, a vertical asymptote?

Answer 5

yes but I dont know how to write that in slop intercept form... x = -1?

Answer 6

\[\Large (y_0+3)(y+3) \ \text{with} \ y_0=-3 \longrightarrow 0 \]

Answer 7

yes, and when you implicit you devide by (y+3) ...

Answer 8

\[ \Large -3x+6=9 \\\Large -3x=3 \\ \Large x=-1\]

Answer 9

that should be it yes.

Answer 10

http://www.wolframalpha.com/input/?i=plot+%28x-2%29%5E2%2B%28y%2B3%29%5E2%3D9+and+x%3D-1 Graph by WolframAlpha.

Answer 11

this is a little tricky to see tammy. Are you doing implicit in class?

Answer 12

Thanks. I am going to try it now

Answer 13

No we are not doing implicit. This is a summer packet

Answer 14

I recommend you to do both, implicit differentiation (for practice and understanding) and you can use the method I have mention above too to check that the answers are equal.

Answer 15

but this is for calc?

Answer 16

Yes pre

Answer 17

*grins* strange course.

Answer 18

Maybe they expect you to write the circle in explicit form and then use regular differentiation. Just make sure that you draw the circle in a xy-plane first so you can check if you need the positive or negative slope of the calculated value.

Answer 19

Ok. Thanks I am going to try it out now and see where it takes me.

Answer 20

are you doing derivatives in class?

Answer 21

Yes. Now one more time can you take me step by step once again i am getting kind of lost.

Answer 22

Theere will always be an asymptote when you are at the center of a circle such as this. So really you did not have to do any work...

Answer 23

just think about the circle. and its asking for the tanget line at the center y point. so its will be straight up and down.

Answer 24

So x=-1

Answer 25

yes

Answer 26

Omg! I am starting to understand now. But why does it take you through so many uneed steps

Answer 27

what does?

Answer 28

this is a special case, they are not all going to be like this. This one you did not have to do any work to now the answer, but most of the time this is not the case.

Answer 29

Ok. Hold on.

Answer 30

Ok. So it said for step 1 identify the center, which is (2,-3). And for step 2 find the slope of the line passing through thee center and the point of tangent, which is 0/3. Now for step 3 it ask find the slope of the tangent line here is where i am lost

Answer 31

Find the slope of the tangent line? You mean the vertical line that has no slope?