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?

- anonymous

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- anonymous

@ganeshie8

- 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

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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## More answers

- welshfella

The circle touches the x -axis so that another point on (x , 0)

- welshfella

- good question - I'm trying to figure that

- anonymous

Can you give a picture?
that is puzzling me

- welshfella

|dw:1432910060214:dw|

- welshfella

lo1 - not a great diagram

- anonymous

I'm sorry I meant two possible cirlces

- welshfella

|dw:1432910234711:dw|

- welshfella

thats worse than the first

- anonymous

:)
Yeah that is amazing.

- ganeshie8

two circles can have the same common tangent, yes ?

- welshfella

yes

- ganeshie8

may be forget about x-axis and look at below diagram
|dw:1432910495765:dw|

- ganeshie8

both circles meet below 3 conditions :
1, 2) pass through two points (intersection)
3) having that blue line as tangent

- welshfella

ah - yes the common tangenytis the x-axis

- anonymous

I got it now
thanks

- 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

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

anyway that for your helping :)

- ganeshie8

hmm the tangent point is not so random, we only have two choices so its still a bit mysterious

- ganeshie8

|dw:1432911213933:dw|

- anonymous

But the result is the same tangent line condition isn't unique as you have said.

- ganeshie8

geometrically how do you know there doesn't exist a third circle that meets the given conditions ?

- ganeshie8

|dw:1432911454842:dw|

- ganeshie8

how do i convince more than two circles are never possible given
1) two points
2) tangent line

- anonymous

the center isn't on locus of chord made by these 2 points

- anonymous

I mean when you bisect it and make a perpendicular which is the locus of the center

- ganeshie8

perpendicular bisector of the chord passes through the center
but again how do you know the 3 centers are not collinear ?

- ganeshie8

|dw:1432911763458:dw|

- 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

BINGO!!!

- ganeshie8

|dw:1432912228871:dw|

- ganeshie8

hmm idk

- welshfella

yes - its a puzzle

- welshfella

|dw:1432912469311:dw|

- welshfella

why can't there be an other larger circle like the above?

- 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

brilliant!

- welshfella

i'm afraid i gotta go and leave this interesting discussion ..

- 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

It is pretty long that way. :p

- 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

looks good to me

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