## anonymous one year ago 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?

1. anonymous

@ganeshie8

2. 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

3. anonymous

Circle is clearly defined by 3 conditions. How can I draw 2 circles which path through two points and tangent to a line!!

4. welshfella

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

5. welshfella

- good question - I'm trying to figure that

6. anonymous

Can you give a picture? that is puzzling me

7. welshfella

|dw:1432910060214:dw|

8. welshfella

lo1 - not a great diagram

9. anonymous

I'm sorry I meant two possible cirlces

10. welshfella

|dw:1432910234711:dw|

11. welshfella

thats worse than the first

12. anonymous

:) Yeah that is amazing.

13. ganeshie8

two circles can have the same common tangent, yes ?

14. welshfella

yes

15. ganeshie8

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

16. ganeshie8

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

17. welshfella

ah - yes the common tangenytis the x-axis

18. anonymous

I got it now thanks

19. ganeshie8

your question about "why 3 conditions are not giving me an unique circle" is very interesting, im still thinking of a better explanation

20. 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))

21. anonymous

anyway that for your helping :)

22. ganeshie8

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

23. ganeshie8

|dw:1432911213933:dw|

24. anonymous

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

25. ganeshie8

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

26. ganeshie8

|dw:1432911454842:dw|

27. ganeshie8

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

28. anonymous

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

29. anonymous

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

30. ganeshie8

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

31. ganeshie8

|dw:1432911763458:dw|

32. 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?

33. ganeshie8

BINGO!!!

34. ganeshie8

|dw:1432912228871:dw|

35. ganeshie8

hmm idk

36. welshfella

yes - its a puzzle

37. welshfella

|dw:1432912469311:dw|

38. welshfella

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

39. 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.

40. ganeshie8

brilliant!

41. welshfella

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

42. myininaya

I know you wanted geometric stuff or whatever but I think I have found two answers with a combination of algebra and calculus.

43. myininaya

It is pretty long that way. :p

44. 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

45. welshfella

looks good to me