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).

- HRazor

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- 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

|dw:1377901398623:dw|

- 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

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

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

what are your ideas?

- anonymous

Interesting question haha its toying with my mind.

- 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

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

im curious if this is designed such that when y=-1, we have the center...

- 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

@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

yes, but without knowing the points of tangency (i cant see them as of yet) its tricky

- 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

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

its a property of tangents to a circle ...

- amistre64

|dw:1377903823180:dw|

- anonymous

Ah! I see, haha I never knew that particular geometric property of circles.

- 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

Good point, however, If it was an optimization problem, we might be able to solve it through derivatives. But its not...

- amistre64

it is a derivative problem due to the tangents

- anonymous

Then ill handle the optimization.

- amistre64

thats "circle" creates a parabola; and we can optimize it to either line: say y-2x+8

- anonymous

Alright..

- amistre64

My times up for today .... ill most likely consider this while laying awake tonight :)

- 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

By the way, My optimal solution is the point: \(\left(\frac{17}{2},9\right)\)

- anonymous

Don't know what good is that though haha

- 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

so i am looking for another way

- anonymous

Which equations?

- 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

Nice method :) I like it! What are the equations?

- 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

here :D

- anonymous

i am not mathematitian so i gave up on solving them :D

- anonymous

by the way sorry for slow writing its my first day here :D

- anonymous

Haha well ill see what I can do and welcome to Openstudy! :)

- anonymous

thank you

- anonymous

oh i got equation only in terms of a and b :D

- anonymous

Ouch that's a nasty bunch of equations!

- anonymous

i should never try to solve this kind of equations i got quadratic equation for a and discriminant isnt nice squre :((

- 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

i got quadratic for a but i think i have some mistake in my arithmatic

- anonymous

i modified this equations a bit i will send new ones if you want

- anonymous

method is theoriticaly posseble but we cant get answer, but nothing else easier comes to my mind

- anonymous

Would this work?
\[(x-3.5)^2+(y+0.5)^2=0.5\]

- amistre64

Ive got 2 circles and a simple enough method developed but this site is not phone froendly

- 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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