## watchmath 5 years ago Limit and circle The problem is attached

1. watchmath

$$C_1$$ is a fixed circle given by $$(x-1)^2+y^2=1$$ and $$C_2$$ is a shriking circle with radius $$r$$. $$P$$ is the point of $$(0,r)$$, $$Q$$ is the upper point of intersection of the two circles,a nd $$R$$ is the point of intersection of the line $$PQ$$ and the $$x$$-axis. WHat happens to $$R$$ as $$C_2$$ shrinks, that is, as $$r\to 0^+$$?

2. anonymous

it moves toward the y axis

3. anonymous

if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

4. anonymous

PLS HELP

5. watchmath

@ishan Do you mean $$\lim_{r\to 0^+}R=0$$? That is not correct!! Try again :D

6. anonymous

the Y(op)-->0 and PR-->0R

7. anonymous

if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

8. myininaya

eventually the shrinking circle will have radius 0 and it will just be a point on the origin and P=Q drawing a line from P to Q would then give a horizontal line on the x axis

9. myininaya

P=Q because C1 and C2 intersect at the origin

10. watchmath

You need to work out this more rigorously. Your intuition can deceive you here.

11. myininaya

the slope of PQR is -r/R (using P=(0,r) and R=(R,0)) so as r goes to zero the slope of PQR are getting closer to zero

12. watchmath

But we are interested in finding R (as r-->0 ). How is the fact that you mention helping us?

13. myininaya

R is the x-intercept of PQ so i'm not seeing how the line PQ isn't approaching the x-axis or y=0

14. myininaya

meaning R would be every point on the x-axis since PQ is that line I don't know its late

15. myininaya

i go to sleep and dream about it tonight k?

16. myininaya

oops this morning its 3:34 am

17. myininaya

well if the slope is approaching 0 and the line isn't moving toward the x-axis, then that would mean the line PQ is getting closer to not having a x-intercept (R) since PQ is a constant not equal to 0

18. watchmath

It is 4:42 here :). I think dumbcow will answer this, right? :D

19. dumbcow

limit as r->0 is 4

20. dumbcow

not 100% but i was able to get R as a function of r

21. watchmath

Yes, then now it is just an exercise of computing limit.

22. dumbcow

$R =\frac{r^{2}}{2-\sqrt{4-r^{2}}}$ use L'hopitals rule

23. dumbcow

first i set the 2 circle equations equal to each other to find Q in terms of r then determined slope with P and Q used equation of a line to find R the x-intercept

24. myininaya

thats right Q is the intersection of the two circles :(

25. dumbcow

btw thanks for the diagram could not have done this without that, im a visual person :)

26. myininaya

ok i guess i see how it can be 4 lol

27. myininaya

what about what i mention? why is it wrong? i mean i know its wrong because it doesn't give the right answer, but what makes it wrong to do

28. myininaya

P=(0,r) R=(R,0) the slope of PQ is r/(-R) this shows as r goes to zero tha PQ is getting closer to being a horizontal line

29. dumbcow

yes slope is getting flatter which pushes the x-intercept farther out so there is just a limit to how far out it can go so you are not wrong on that point

30. myininaya

but it never totally goes flat okay i get it

31. myininaya

meaning there is a x-intercept

32. dumbcow

right P approaches the origin and PQ approaches the x-axis but dont ever get there :)

33. myininaya

because we never get to the origin lol

34. myininaya

thanks cow

35. dumbcow

36. myininaya

watchmath give me another i want to prove i can think like a pro

37. myininaya

don't do it now wait til tomorrow let me sleep if you give it to me now then i will never sleep

38. anonymous

ok the fixed circle EQ is (x-1)^2 +y^2=1...the other circle is at the center with (h,k)=(0,0) therefore the EQ is x^2 + y^2 =r^2 are you geting it watch?

39. anonymous

now in this two EQ, you can compute the intersection PQ and get its X and Y

40. anonymous

now once you get the x and y, you can get its hypotenuse which is the PQ..then from there you can set the limit as r-->0

41. anonymous

now i dont know if Q is fixed there at that point, or changed position with the other circle, as r--->0