HRazor
  • HRazor
Derive the equation of the circle tangent to the lines x - 2y + 4 = 0 and 2x - y - 8 = 0 and passes through the point (4, -1).
Mathematics
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
|dw:1377901291946:dw| well, the center of the circle will be along the line that bisects the angle formed at the point they cross
amistre64
  • amistre64
|dw:1377901398623:dw|
amistre64
  • amistre64
we have the point of contact at (20/3, 16/3) so this is going to be a nice and fractioned playground ....

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amistre64
  • amistre64
so about 7,5 .. and the point 4.-1 is to be included |dw:1377901631739:dw| might be good to know where that stray point is relative to the lines ...
amistre64
  • amistre64
x - 2y = -4 -1 - 2(4) = -4 -9 < -4, so its below that line 2x - y - 8 = 0 -2 - 4 = 8 and below that line .... |dw:1377901810652:dw| so something of this kind of orientation ... hmm
amistre64
  • amistre64
what are your ideas?
anonymous
  • anonymous
Interesting question haha its toying with my mind.
amistre64
  • amistre64
(x-20/3)^2+(2x-8-16/3)^2 = 20; x = 14/3,. y = 4/3 (x-20/3)^2+(x/2+2-16/3)^2 = 20; x = 32/3, y = 22/3 this gives me 2 points to "midpoint" (23/3, 13/3) so we have a line for the center to ride along .... between (20/3, 16/3) and (23/3, 13/3) (20/3, 16/3) -(23/3, 13/3) ------------- -3/3 , 3/3 ... a slope of -1 eh ... y = -(x-20/3)+16/3 y = -x+12 so we have a frame work now
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=x-2y%3D-4%2C+2x-y%3D8%2C+y%3D-1%2C+x%3D4%2C+y+%3D+-x%2B12
amistre64
  • amistre64
im curious if this is designed such that when y=-1, we have the center...
amistre64
  • amistre64
y = -x+12 -1 = -x + 12, x = 13 (13,-1) (4,-1) is a distance (radius) of 9 .... just as a trial and error
anonymous
  • anonymous
@amistre64 Would it be fair to say that the circle is centred at the intersection between the lines that perpendicularly intersect each of the tangent lines at the tangent point?
amistre64
  • amistre64
yes, but without knowing the points of tangency (i cant see them as of yet) its tricky
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=%28x-13%29%5E2%2B%28y%2B1%29%5E2+%3D+81%2C+x-2y%3D-4%2C+2x-y%3D8%2C+y+%3D+-x%2B12 my attempt to "guess" the center makes it a little to big
anonymous
  • anonymous
Yeah I just realized that now!...However, how do you know that the center of the circle will be along the line that bisects the angle formed at the point they cross?
amistre64
  • amistre64
its a property of tangents to a circle ...
amistre64
  • amistre64
|dw:1377903823180:dw|
anonymous
  • anonymous
Ah! I see, haha I never knew that particular geometric property of circles.
amistre64
  • amistre64
im thinking at the moment, that for some center: (x,-x+12), and some radius (r), that satisfies the given points (4,-1) ... (4-x)^2 + (-1-(-x+12))^2 = r^2
anonymous
  • anonymous
Good point, however, If it was an optimization problem, we might be able to solve it through derivatives. But its not...
amistre64
  • amistre64
it is a derivative problem due to the tangents
anonymous
  • anonymous
Then ill handle the optimization.
amistre64
  • amistre64
thats "circle" creates a parabola; and we can optimize it to either line: say y-2x+8
anonymous
  • anonymous
Alright..
amistre64
  • amistre64
My times up for today .... ill most likely consider this while laying awake tonight :)
anonymous
  • anonymous
Yeah so will I. Im trying to use lagrangian multipliers but if that doesn't work, ill see what I can do for like an hour and probably end up think about it...hopefully not in my dreams haha
anonymous
  • anonymous
By the way, My optimal solution is the point: \(\left(\frac{17}{2},9\right)\)
anonymous
  • anonymous
Don't know what good is that though haha
anonymous
  • anonymous
i got 3 equations in terms of a, b and r (x-a)^2+(y-b)^2=r^2 but they are too ugly :D
anonymous
  • anonymous
so i am looking for another way
anonymous
  • anonymous
Which equations?
anonymous
  • anonymous
i wrote 2 equations for tangent lines of circle \[\frac{ dy }{ dx }=-\frac{ x-a }{ y-b }\] so i found touching points of 2 given tangent lines to circle in terms of a and b. this points are also points on circle so they must satisfy equation of circle. this is two equations and 3rd equation is in terms of point
anonymous
  • anonymous
Nice method :) I like it! What are the equations?
anonymous
  • anonymous
\[\left( 4-a \right)^{2}+\left( b+1 \right)^{2}=r ^{2}\]\[\left( 0.4b-0.2a-0.8 \right)^{2}+\left( 0.4a-0.8b+1.6 \right)^{2}=r ^{2}\]\[\left( 0.4b-0.8a+3.2 \right)^{2}+\left( 0.4a-0.2b-1.6\right)^{2}=r ^{2}\]
anonymous
  • anonymous
here :D
anonymous
  • anonymous
i am not mathematitian so i gave up on solving them :D
anonymous
  • anonymous
by the way sorry for slow writing its my first day here :D
anonymous
  • anonymous
Haha well ill see what I can do and welcome to Openstudy! :)
anonymous
  • anonymous
thank you
anonymous
  • anonymous
oh i got equation only in terms of a and b :D
anonymous
  • anonymous
Ouch that's a nasty bunch of equations!
anonymous
  • anonymous
i should never try to solve this kind of equations i got quadratic equation for a and discriminant isnt nice squre :((
anonymous
  • anonymous
The only relation I was able to reach was: \[a=\frac{8\pm\sqrt{64+0.72b^2-7.68b-11.52}}{0.6}\]
anonymous
  • anonymous
i got quadratic for a but i think i have some mistake in my arithmatic
anonymous
  • anonymous
i modified this equations a bit i will send new ones if you want
anonymous
  • anonymous
method is theoriticaly posseble but we cant get answer, but nothing else easier comes to my mind
anonymous
  • anonymous
Would this work? \[(x-3.5)^2+(y+0.5)^2=0.5\]
amistre64
  • amistre64
Ive got 2 circles and a simple enough method developed but this site is not phone froendly
amistre64
  • amistre64
y=2x-8 has a perp: -1/2 (x-a) +(2a-8) this intersects the center line at the point: (40-5a , 5a-28) when the distance between this point and the points (4,-1) and (a,2a-8) we have a quadratic in a. solving gives me a=1 or 5. so at the point 15,-3; radius 125. and at 35,-23 r 1445 produce the required circles

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