# intersecting circles

• Mar 13th 2009, 11:01 AM
FRANTICSTAR
intersecting circles
Hi I'd like to know how I find out where 2 circles would intersect

I can show the following example:

http://dev.franticstar.co.uk/trig/trig.jpg

So far I have:

Left circle:
x = 250 + 200 * cos(a1)
y = 350 + 200 * sin(a1)

where a1 has 2 values where the angles where the circles intersect

and the right circle:
x = 400 + 200 * cos(a2)
y = 250 + 200 * sin(a2)

where a2 has 2 values where the angles where the circles intersect

from that I have:

250 + 200 * cos(a1) = 400 + 200 * cos(a2)
350 + 200 * sin(a1) = 250 + 200 * sin(a2)

Which is where I get stuck.
• Mar 13th 2009, 11:35 AM
running-gag
Hi

Maybe it is more simple with Cartesian equations

The first circle equation is $(x-250)^2+(y-350)^2=200^2$
The second circle equation is $(x-400)^2+(y-250)^2=200^2$

A point on the two circles satisfies both equations

Subtracting gives
$300x-200y+350^2-400^2=0$

$y=\frac32\:x-\frac{375}{2}$

Substituting in equation 2 gives a quadratic equation whose solutions are approximately
$x_1 = 225.97$ => $y_1 = 151.45$
$x_2 = 424.03$ => $y_2 = 448.55$
• Mar 13th 2009, 10:20 PM
FRANTICSTAR
Hi running-gag,

This looks great, I've never heard of Cartesian equations, i'm a little confused how you get to:

300x-200y+350^2-400^2=0

and

y = 3/2x - 375/2
• Mar 13th 2009, 11:22 PM
Reckoner
Quote:

Originally Posted by FRANTICSTAR
This looks great, I've never heard of Cartesian equations

You have learned about parametric equations, but not Cartesian (rectangular) ones? I find that a bit hard to believe.

Quote:

300x-200y+350^2-400^2=0
As running-gag said, you subtract:

$\begin{tabular}{cc}
&(x-250)^2+(y-350)^2=200^2\\
-&(x-400)^2+(y-250)^2=200^2\\\hline
\end{tabular}$

which gives

$(x-250)^2+(y-350)^2-\left[(x-400)^2+(y-250)^2\right]=200^2-200^2$

$\Rightarrow(x-250)^2+(y-350)^2-(x-400)^2-(y-250)^2=0$

$\Rightarrow(x^2-500x+250^2)+(y^2-700y+350^2)-(x^2-800x+400^2)-(y^2-500y+250^2)=0$

$\Rightarrow300x-200y+350^2-400^2=0.$

Quote:

y = 3/2x - 375/2
Continuing from the above,

$300x-200y+350^2-400^2=0$

$\Rightarrow3x-2y+35^2-40^2=0$

$\Rightarrow y=\frac{3x+35^2-40^2}2$

$\Rightarrow y=\frac32x+\frac{35^2-40^2}2$

$\Rightarrow y=\frac32x+\frac{25(7^2-8^2)}2$

$\Rightarrow y=\frac32x+\frac{25(-15)}2$

$\Rightarrow y=\frac32x-\frac{375}2.$
• Mar 14th 2009, 12:22 AM
FRANTICSTAR
Cool I think that's starting to make some sense!

I didn't know they were called parametric equations, I learnt those equations from programming my amastrad when I was 11.