DLS
  • DLS
The equation of circle which passes through (1,0),(0,1) & has its Radius as smal as possible is?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ParthKohli
  • ParthKohli
There's only one unique circle passing through (1,0) and (0,1) right?
anonymous
  • anonymous
No.
DLS
  • DLS
i dont think so

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ParthKohli
  • ParthKohli
Thanks for clearing my doubts. That's what I was trying out
ParthKohli
  • ParthKohli
May I have a counterexample BTW?
DLS
  • DLS
radius is your choice
anonymous
  • anonymous
|dw:1357382125336:dw|
ParthKohli
  • ParthKohli
Thanks.
ParthKohli
  • ParthKohli
Two circles that intersect, right. =)
anonymous
  • anonymous
the smallest circle is the one that has the line connecting the two points as the diameter.
DLS
  • DLS
wait what did you say :?
DLS
  • DLS
and why are there 2 circles
ParthKohli
  • ParthKohli
Radius of \(1\)? |dw:1357382192536:dw|
ParthKohli
  • ParthKohli
That's just the unit circle.
anonymous
  • anonymous
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
  • ParthKohli
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|
anonymous
  • anonymous
why all this?
ParthKohli
  • ParthKohli
\[r^2 = h^2 + (k - 1)^2 = (h - 1)^2 + k^2\]
DLS
  • DLS
I just want to know why should they be diameterically opp..to satisfy the eq.nothing else :/
ParthKohli
  • ParthKohli
\[h^2 + k^2 -2k+1=h^2 - 2h + 1 + k^2\]\[-2h + 1 = -2k +1 \iff h =k\]So \((0,0)\)?
ParthKohli
  • ParthKohli
\((0,0)\) is the center.
DLS
  • DLS
why u no tell jitna pucha :(
ParthKohli
  • ParthKohli
1 answer
ParthKohli
  • ParthKohli
\[x^2 + y^2 = 1\]
ParthKohli
  • ParthKohli
Pakka yahi hai answer.
DLS
  • DLS
lol no way
ParthKohli
  • ParthKohli
Why this Kolaveri
anonymous
  • anonymous
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
  • DLS
not so clear but yeah somewhat..
ParthKohli
  • ParthKohli
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
  • DLS
how can (1,0) and (0,1) be diameterically opp but :O
ParthKohli
  • ParthKohli
I didn't say they were diametrically opposite =/
DLS
  • DLS
:S
anonymous
  • anonymous
but i did.
anonymous
  • anonymous
they have to be.
ParthKohli
  • ParthKohli
They are diametrically perpendicular...
DLS
  • DLS
how on eart h:/
anonymous
  • anonymous
No parth. I don't even know what that means... But i'll draw on paper and show you what I mean.
ParthKohli
  • ParthKohli
|dw:1357383885601:dw|
DLS
  • DLS
kya yaa dimaag ka dahi mat karo :P aur bhi q hai!!
anonymous
  • anonymous
that's absolutely wrong. I can draw a smaller circle!|dw:1357384070442:dw| you see the point!? this right here the smallest circle.
ParthKohli
  • ParthKohli
Oh, never thought of that =P
DLS
  • DLS
OH ! !
ParthKohli
  • ParthKohli
So then it's\[(x - 0.5)^2 + (y - 0.5)^2 = 0.25\]?
DLS
  • DLS
no
ParthKohli
  • ParthKohli
So that's the trick. You make them diametrically opposite
ParthKohli
  • ParthKohli
@rajathsbhat Yaar chaa gaye =P
DLS
  • DLS
(x-x1)(x-x2)+(y-y1)(y-y2)=0 hoti hai equation diameterically op points ki :D
anonymous
  • anonymous
that's right^
DLS
  • DLS
aaree yaar aage bado :P
ParthKohli
  • ParthKohli
But we want the equation of the circle!
DLS
  • DLS
yeah wo circle ki he hai
ParthKohli
  • ParthKohli
Achcha okay! Point noted

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