Limit and circle
The problem is attached

- watchmath

Limit and circle
The problem is attached

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

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- anonymous

it moves toward the y axis

- anonymous

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

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## More answers

- anonymous

PLS HELP

- watchmath

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

- anonymous

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

- anonymous

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

- 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

- myininaya

P=Q because C1 and C2 intersect at the origin

- watchmath

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

- 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

- watchmath

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

- 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

- myininaya

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

- myininaya

i go to sleep and dream about it tonight k?

- myininaya

oops this morning its 3:34 am

- 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

- watchmath

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

- dumbcow

limit as r->0 is 4

- dumbcow

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

- watchmath

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

- dumbcow

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

- 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

- myininaya

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

- dumbcow

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

- myininaya

ok i guess i see how it can be 4 lol

- 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

- 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

- 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

- myininaya

but it never totally goes flat
okay i get it

- myininaya

meaning there is a x-intercept

- dumbcow

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

- myininaya

because we never get to the origin lol

- myininaya

thanks cow

- dumbcow

your welcome
watchmath did that answer all your questions

- myininaya

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

- myininaya

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

- 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?

- anonymous

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

- 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

- anonymous

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

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