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anonymous
 one year ago
Find circle which is tangent to xaxis 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?
anonymous
 one year ago
Find circle which is tangent to xaxis 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?

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.2general 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
 one year ago
Best ResponseYou've already chosen the best response.0Circle is clearly defined by 3 conditions. How can I draw 2 circles which path through two points and tangent to a line!!

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2The circle touches the x axis so that another point on (x , 0)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2 good question  I'm trying to figure that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you give a picture? that is puzzling me

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2dw:1432910060214:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2lo1  not a great diagram

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry I meant two possible cirlces

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2dw:1432910234711:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2thats worse than the first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:) Yeah that is amazing.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1two circles can have the same common tangent, yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1may be forget about xaxis and look at below diagram dw:1432910495765:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1both circles meet below 3 conditions : 1, 2) pass through two points (intersection) 3) having that blue line as tangent

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2ah  yes the common tangenytis the xaxis

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1your question about "why 3 conditions are not giving me an unique circle" is very interesting, im still thinking of a better explanation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0anyway that for your helping :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1hmm the tangent point is not so random, we only have two choices so its still a bit mysterious

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432911213933:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But the result is the same tangent line condition isn't unique as you have said.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1geometrically how do you know there doesn't exist a third circle that meets the given conditions ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432911454842:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1how do i convince more than two circles are never possible given 1) two points 2) tangent line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the center isn't on locus of chord made by these 2 points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean when you bisect it and make a perpendicular which is the locus of the center

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1perpendicular bisector of the chord passes through the center but again how do you know the 3 centers are not collinear ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432911763458:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432912228871:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2dw:1432912469311:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2why can't there be an other larger circle like the above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.2i'm afraid i gotta go and leave this interesting discussion ..

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I know you wanted geometric stuff or whatever but I think I have found two answers with a combination of algebra and calculus.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2It is pretty long that way. :p

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ our circle has points: } (1,2);(3,4);(a,0) \\ \text{ we have horizontal tangent at } (a,0) \\ (ah)^2+(0k)^2=r^2 \\ (1h)^2+(k+2)^2=r^2 \\ (3h)^2+(k+4)^2=r^2 \\ y'=\frac{hx}{yk} \\ y'_{x=a}=0=\frac{ha}{0k} \implies h=a \\ (aa)^2+k^2=r^2 \text{ so } k=r \text{ or } k=r \\ \text{ so going with the } k=r \text{ thing we have } \\ (1h)^2+(r+2)^2=r^2 \\ (3h)^2+(r+4)^2=r^2 \\ (1h)^2+4r+4=0 \\ (3h)^2+8r+16=0 \\ 2(1h)^2+8=(3h)^2+16 \\ h^2+2h15=0 \\ (h+5)(h3)=0 \\ h=5 \text{ or } h=3 \] so pluggin some things back in we can find r then k \[h=5 \\ a=5 \\ (1+5)^2+4r+4=0 \\ r=10 \\ k=10,10\] \[h=3 \\ a=3 \\ (13)^2+4r+4=0 \\ r=2 \\ k=2 ,2\] 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
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