My Math Forum Define circle when:

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 April 22nd, 2014, 07:40 AM #1 Newbie   Joined: Apr 2014 From: Czech Republic Posts: 4 Thanks: 0 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: $\displaystyle (x-m)^2 + (y-n)^2 = r^2$
 April 22nd, 2014, 07:50 AM #2 Senior Member   Joined: Apr 2014 From: UK Posts: 903 Thanks: 331 I'd draw the 3 lines then put the circle on it and work out what it is from that.
 April 22nd, 2014, 07:56 AM #3 Newbie   Joined: Apr 2014 From: Czech Republic Posts: 4 Thanks: 0 Ye, well, thing is I'm not allowed to draw it.
 April 22nd, 2014, 08:26 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra Look carefully at the equations. What do you notice about the two lines a and b? Thanks from Hakundin
 April 22nd, 2014, 09:02 AM #5 Newbie   Joined: Apr 2014 From: Czech Republic Posts: 4 Thanks: 0 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.
 April 22nd, 2014, 09:14 AM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond $\displaystyle \frac43x+\frac{10}{3}\,=\,-\frac34x\Rightarrow(x,y)=\left(-\frac85,\frac65\right)$ $\displaystyle \frac43x-10\,=\,-\frac34x\Rightarrow(x,y)=\left(\frac{24}{5},-\frac{18}{5}\right)$ $\displaystyle \Rightarrow r=4$ Distance from a point to a line - Wikipedia, the free encyclopedia $\displaystyle \frac{|4x-3(-2x+10)+10|}{5}=4\Rightarrow x=4,y=2$ $\displaystyle (x-4)^2+(y-2)^2=16$ Thanks from Hakundin
April 22nd, 2014, 09:40 AM   #7
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Quote:
 Originally Posted by greg1313 $\displaystyle \Rightarrow r=4$

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.

April 22nd, 2014, 10:01 AM   #8
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Quote:
 Originally Posted by Hakundin $\displaystyle v(A, b)= |(4*5)+((10*(-3))-(-30)| ÷ (\sqrt{25}))$
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.

Last edited by greg1313; April 22nd, 2014 at 11:38 AM.

 April 22nd, 2014, 02:33 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 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). Thanks from greg1313

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