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April 22nd, 2014, 07:40 AM   #1
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Define circle when:

Heya, got a little problem which I can't solve for a while now.

'Define a circle which has it's center on a straight line "p" and which touches straight lines "a" and "b".

$\displaystyle p: 2x + y - 10 = 0$

$\displaystyle a: 4x - 3y + 10 = 0$
$\displaystyle b: 4x - 3y - 30 = 0$

I've got absolutely zero idea even how to start.
Only thing I know is that circle if defined by:
$\displaystyle (x-m)^2 + (y-n)^2 = r^2 $
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April 22nd, 2014, 07:50 AM   #2
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I'd draw the 3 lines then put the circle on it and work out what it is from that.
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April 22nd, 2014, 07:56 AM   #3
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Ye, well, thing is I'm not allowed to draw it.
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April 22nd, 2014, 08:26 AM   #4
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Look carefully at the equations. What do you notice about the two lines a and b?
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April 22nd, 2014, 09:02 AM   #5
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They are parallel? One intercects y axis at [0; 10] and another one at [0; -30].*
So if I find distance between them, I'll get my 2*r . Now I just need to find out where the circle's center is... somehow.

*actually, maybe they don't intercect y axis at these coordinates. Nevertheless I think I found out how to get d distance between these lines.

Last edited by Hakundin; April 22nd, 2014 at 09:14 AM.
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April 22nd, 2014, 09:14 AM   #6
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$\displaystyle \frac43x+\frac{10}{3}\,=\,-\frac34x\Rightarrow(x,y)=\left(-\frac85,\frac65\right)$

$\displaystyle \frac43x-10\,=\,-\frac34x\Rightarrow(x,y)=\left(\frac{24}{5},-\frac{18}{5}\right)$

$\displaystyle \Rightarrow r=4$

Distance from a point to a line - Wikipedia, the free encyclopedia

$\displaystyle \frac{|4x-3(-2x+10)+10|}{5}=4\Rightarrow x=4,y=2$

$\displaystyle (x-4)^2+(y-2)^2=16$
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April 22nd, 2014, 09:40 AM   #7
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle \Rightarrow r=4$

For some reason, using the same technique (distance from a point to a line) I've gotten different $\displaystyle r$.

I chose point A $\displaystyle A[5; 10] $ on line $\displaystyle a $ and calculate dinstace:

$\displaystyle v(A, b)= |(4*5)+((10*(-3))-(-30)| ÷ (\sqrt{25})) $
$\displaystyle v(A, b) = 4 = d$
$\displaystyle r= d÷2$
$\displaystyle r=2$


and I don't quite understand how exactly did you get $\displaystyle x=4$ and $\displaystyle y=2$

Last edited by Hakundin; April 22nd, 2014 at 09:44 AM.
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April 22nd, 2014, 10:01 AM   #8
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Quote:
Originally Posted by Hakundin View Post
$\displaystyle v(A, b)= |(4*5)+((10*(-3))-(-30)| ÷ (\sqrt{25})) $
That should be $\displaystyle v(A, b)= |4*5+10*(-3)+(-30)| ÷ \sqrt{25} $

Quote:
and I don't quite understand how exactly did you get x=4 and y=2
I used x and y = -2x + 10 as the point for the distance from a point to a line formula to solve for x. Then y = -2(4) + 10 = 2.
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Last edited by greg1313; April 22nd, 2014 at 11:38 AM.
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April 22nd, 2014, 02:33 PM   #9
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Once you've noticed that the two lines are parallel, it is easy too write the equation of a thirrd parallel line on which the centre must lie. But we have another line on which the centre must also lie, so the centre of the circle is at the point of intersection of those two lines.

The diameter is, as you said, the shortest distance between a and b, but now we've found the centre of the circle, it may be easier to find the point at which a (or b) has only a single point of intersection with the circle (or a repeated point of intersection).
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