anonymous
  • anonymous
Find circle which is tangent to x-axis and path through points (1,-2) and (3,-4). I worked it through analytic geometry and it gives two possible circles!!! and I can not figure how 3 conditions for a circle gives 2 answers or how to draw it geometrically?
Mathematics
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anonymous
  • anonymous
Find circle which is tangent to x-axis and path through points (1,-2) and (3,-4). I worked it through analytic geometry and it gives two possible circles!!! and I can not figure how 3 conditions for a circle gives 2 answers or how to draw it geometrically?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
welshfella
  • welshfella
general equation is (x - a)^2 + ( y - b)^2 = r^2 you can create 2 equations in a , b and r by substituting the 2 given points
anonymous
  • anonymous
Circle is clearly defined by 3 conditions. How can I draw 2 circles which path through two points and tangent to a line!!

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welshfella
  • welshfella
The circle touches the x -axis so that another point on (x , 0)
welshfella
  • welshfella
- good question - I'm trying to figure that
anonymous
  • anonymous
Can you give a picture? that is puzzling me
welshfella
  • welshfella
|dw:1432910060214:dw|
welshfella
  • welshfella
lo1 - not a great diagram
anonymous
  • anonymous
I'm sorry I meant two possible cirlces
welshfella
  • welshfella
|dw:1432910234711:dw|
welshfella
  • welshfella
thats worse than the first
anonymous
  • anonymous
:) Yeah that is amazing.
ganeshie8
  • ganeshie8
two circles can have the same common tangent, yes ?
welshfella
  • welshfella
yes
ganeshie8
  • ganeshie8
may be forget about x-axis and look at below diagram |dw:1432910495765:dw|
ganeshie8
  • ganeshie8
both circles meet below 3 conditions : 1, 2) pass through two points (intersection) 3) having that blue line as tangent
welshfella
  • welshfella
ah - yes the common tangenytis the x-axis
anonymous
  • anonymous
I got it now thanks
ganeshie8
  • ganeshie8
your question about "why 3 conditions are not giving me an unique circle" is very interesting, im still thinking of a better explanation
anonymous
  • anonymous
I don't know. In such cases I just change that thinking. I think the tangent line condition isn't tough enough ((unless the tangent point is given))
anonymous
  • anonymous
anyway that for your helping :)
ganeshie8
  • ganeshie8
hmm the tangent point is not so random, we only have two choices so its still a bit mysterious
ganeshie8
  • ganeshie8
|dw:1432911213933:dw|
anonymous
  • anonymous
But the result is the same tangent line condition isn't unique as you have said.
ganeshie8
  • ganeshie8
geometrically how do you know there doesn't exist a third circle that meets the given conditions ?
ganeshie8
  • ganeshie8
|dw:1432911454842:dw|
ganeshie8
  • ganeshie8
how do i convince more than two circles are never possible given 1) two points 2) tangent line
anonymous
  • anonymous
the center isn't on locus of chord made by these 2 points
anonymous
  • anonymous
I mean when you bisect it and make a perpendicular which is the locus of the center
ganeshie8
  • ganeshie8
perpendicular bisector of the chord passes through the center but again how do you know the 3 centers are not collinear ?
ganeshie8
  • ganeshie8
|dw:1432911763458:dw|
welshfella
  • welshfella
the line joining the points of contact to the centers of the circles are perpendicular to the tangent - can that be used to answer the question?
ganeshie8
  • ganeshie8
BINGO!!!
ganeshie8
  • ganeshie8
|dw:1432912228871:dw|
ganeshie8
  • ganeshie8
hmm idk
welshfella
  • welshfella
yes - its a puzzle
welshfella
  • welshfella
|dw:1432912469311:dw|
welshfella
  • welshfella
why can't there be an other larger circle like the above?
anonymous
  • anonymous
for any triangle it has only one unique circle paths through its vertices |dw:1432912656462:dw| so if we moved the point along the locus the lengths won't be equal and if we shifted it up or down, the the distance to 2pionts won't be equal.
ganeshie8
  • ganeshie8
brilliant!
welshfella
  • welshfella
i'm afraid i gotta go and leave this interesting discussion ..
myininaya
  • myininaya
I know you wanted geometric stuff or whatever but I think I have found two answers with a combination of algebra and calculus.
myininaya
  • myininaya
It is pretty long that way. :p
myininaya
  • myininaya
\[\text{ our circle has points: } (1,-2);(3,-4);(a,0) \\ \text{ we have horizontal tangent at } (a,0) \\ (a-h)^2+(0-k)^2=r^2 \\ (1-h)^2+(k+2)^2=r^2 \\ (3-h)^2+(k+4)^2=r^2 \\ y'=\frac{h-x}{y-k} \\ y'|_{x=a}=0=\frac{h-a}{0-k} \implies h=a \\ (a-a)^2+k^2=r^2 \text{ so } k=-r \text{ or } k=r \\ \text{ so going with the } k=r \text{ thing we have } \\ (1-h)^2+(r+2)^2=r^2 \\ (3-h)^2+(r+4)^2=r^2 \\ (1-h)^2+4r+4=0 \\ (3-h)^2+8r+16=0 \\ 2(1-h)^2+8=(3-h)^2+16 \\ h^2+2h-15=0 \\ (h+5)(h-3)=0 \\ h=-5 \text{ or } h=3 \] so pluggin some things back in we can find r then k \[h=-5 \\ a=-5 \\ (1+5)^2+4r+4=0 \\ r=10 \\ k=10,-10\] \[h=3 \\ a=3 \\ (1-3)^2+4r+4=0 \\ r=2 \\ k=2 ,-2\] some of these solutions need to be checked you will see only 2 of the 4 will work and like i said I know this isn't the approach you wanted but I thought it would be nice to try another approach
welshfella
  • welshfella
looks good to me

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