Question

Limit and circle
The problem is attached

Answer 1

\(C_1\) 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 \(r\to 0^+\)?

Answer 2

it moves toward the y axis

Answer 3

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

Answer 4

PLS HELP

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

P=Q because C1 and C2 intersect at the origin

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

i go to sleep and dream about it tonight k?

Answer 16

oops this morning its 3:34 am

Answer 17

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

Answer 18

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

Answer 19

limit as r->0 is 4

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

ok i guess i see how it can be 4 lol

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

but it never totally goes flat
okay i get it

Answer 31

meaning there is a x-intercept

Answer 32

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

Answer 33

because we never get to the origin lol

Answer 34

thanks cow

Answer 35

your welcome
watchmath did that answer all your questions

Answer 36

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

Answer 37

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

Answer 38

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?

Answer 39

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

Answer 40

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

Answer 41

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