DLS
The equation of circle which passes through (1,0),(0,1) & has its Radius as smal as possible is?
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ParthKohli
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There's only one unique circle passing through (1,0) and (0,1) right?
rajathsbhat
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No.
DLS
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i dont think so
ParthKohli
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Thanks for clearing my doubts. That's what I was trying out
ParthKohli
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May I have a counterexample BTW?
DLS
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radius is your choice
rajathsbhat
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|dw:1357382125336:dw|
ParthKohli
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Thanks.
ParthKohli
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Two circles that intersect, right. =)
rajathsbhat
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the smallest circle is the one that has the line connecting the two points as the diameter.
DLS
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wait what did you say :?
DLS
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and why are there 2 circles
ParthKohli
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Radius of \(1\)?
|dw:1357382192536:dw|
ParthKohli
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That's just the unit circle.
rajathsbhat
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all other circles have this line as their chord which means that their diameters are longer that this line. An extreme example:|dw:1357382299966:dw|
ParthKohli
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If you draw a line segment, you'd get a segment with length \(\sqrt 2\). That is the smallest segment (and the only) possible. Now you get a triangle. Draw a circle around that triangle and that would be \(x^2 + y^2 = 1\).|dw:1357382531256:dw|
rajathsbhat
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why all this?
ParthKohli
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\[r^2 = h^2 + (k - 1)^2 = (h - 1)^2 + k^2\]
DLS
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I just want to know why should they be diameterically opp..to satisfy the eq.nothing else :/
ParthKohli
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\[h^2 + k^2 -2k+1=h^2 - 2h + 1 + k^2\]\[-2h + 1 = -2k +1 \iff h =k\]So \((0,0)\)?
ParthKohli
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\((0,0)\) is the center.
DLS
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why u no tell jitna pucha :(
ParthKohli
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1 answer
ParthKohli
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\[x^2 + y^2 = 1\]
ParthKohli
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Pakka yahi hai answer.
DLS
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lol no way
ParthKohli
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Why this Kolaveri
rajathsbhat
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the concept is pretty straightforward. Look at this diagram|dw:1357382709667:dw|
The line you see is the line connecting (1,0) and (0,1)
And i've drawn two circles that pass though the two points.
The line drawn is a chord for the bigger circle which means that it's diameter is bigger that the line, ok?
But look at the other circle which has the line as the diameter. It's obviously smalller than the other circle but also smaller than any other circle that can be drawn though those two points (again because chords are smaller than diameters)|dw:1357383015575:dw|
So, the circle you're looking for is the circle that has that line for it's diameter. Simple.
DLS
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not so clear but yeah somewhat..
ParthKohli
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another way to think is that you're looking for the smallest right triangle \(ABC\) where \(\angle B\) is the right angle as well as the center of the circle and the circle passes through \(A\) and \(C\).
DLS
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how can (1,0) and (0,1) be diameterically opp but :O
ParthKohli
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I didn't say they were diametrically opposite =/
DLS
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:S
rajathsbhat
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but i did.
rajathsbhat
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they have to be.
ParthKohli
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They are diametrically perpendicular...
DLS
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how on eart h:/
rajathsbhat
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No parth. I don't even know what that means...
But i'll draw on paper and show you what I mean.
ParthKohli
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|dw:1357383885601:dw|
DLS
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kya yaa dimaag ka dahi mat karo :P
aur bhi q hai!!
rajathsbhat
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that's absolutely wrong. I can draw a smaller circle!|dw:1357384070442:dw|
you see the point!? this right here the smallest circle.
ParthKohli
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Oh, never thought of that =P
DLS
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OH ! !
ParthKohli
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So then it's\[(x - 0.5)^2 + (y - 0.5)^2 = 0.25\]?
DLS
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no
ParthKohli
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So that's the trick. You make them diametrically opposite
ParthKohli
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@rajathsbhat Yaar chaa gaye =P
DLS
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(x-x1)(x-x2)+(y-y1)(y-y2)=0
hoti hai equation diameterically op points ki :D
rajathsbhat
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that's right^
DLS
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aaree yaar aage bado :P
ParthKohli
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But we want the equation of the circle!
DLS
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yeah wo circle ki he hai
ParthKohli
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Achcha okay! Point noted