Limit and circle The problem is attached

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Limit and circle The problem is attached

Mathematics
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\(C_1\) 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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it moves toward the y axis
if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

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@ishan Do you mean \(\lim_{r\to 0^+}R=0\)? That is not correct!! Try again :D
the Y(op)-->0 and PR-->0R
if a^3 = b^3 + c^3 + d^3 , then least value of a is ?
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
P=Q because C1 and C2 intersect at the origin
You need to work out this more rigorously. Your intuition can deceive you here.
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
But we are interested in finding R (as r-->0 ). How is the fact that you mention helping us?
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
meaning R would be every point on the x-axis since PQ is that line I don't know its late
i go to sleep and dream about it tonight k?
oops this morning its 3:34 am
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
It is 4:42 here :). I think dumbcow will answer this, right? :D
limit as r->0 is 4
not 100% but i was able to get R as a function of r
Yes, then now it is just an exercise of computing limit.
\[R =\frac{r^{2}}{2-\sqrt{4-r^{2}}}\] use L'hopitals rule
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
thats right Q is the intersection of the two circles :(
btw thanks for the diagram could not have done this without that, im a visual person :)
ok i guess i see how it can be 4 lol
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
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
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
but it never totally goes flat okay i get it
meaning there is a x-intercept
right P approaches the origin and PQ approaches the x-axis but dont ever get there :)
because we never get to the origin lol
thanks cow
your welcome watchmath did that answer all your questions
watchmath give me another i want to prove i can think like a pro
don't do it now wait til tomorrow let me sleep if you give it to me now then i will never sleep
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?
now in this two EQ, you can compute the intersection PQ and get its X and Y
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
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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