Circle intersection

I am trying to find out if these two circles intersect with eachother

(x-2)2 +(y+2)2 = 1

how do I do this?
tango2009Asked:
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ozoConnect With a Mentor Commented:
The first two equations are circles.
I'm not sure what the next four quantities represent.
Nor what the line in the last equation represents.
Whatever you are doing is unnecessary to check whether they intersect or not.
For that, it is sufficient to determine the radius of each circle and the distance between their centres.
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d-glitchCommented:
That's only one circle.
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d-glitchCommented:
The way to check if two circles intersect is to look at the distance between their centers
and the sum of their radii.
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d-glitchCommented:
Do you know how to find the center and radius of this circle:

        (x-2)² +(y+2)² = 1
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phoffricCommented:
This has nice pictures to help visualize:
      http://local.wasp.uwa.edu.au/~pbourke/geometry/2circle/

It not only addresses your question, but also identifies the points of intersection.

There is even a link at the top that has a C source code example by Tim Voght.
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tango2009Author Commented:
sorry made a mistake in posting its these two circles

(x-2)2 +(y+2)2 = 1
x2 +y2 = 4

thanks for your reply phroffic I will have a go using your link
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phoffricCommented:
The previous link was for good pictures and some c-code.
Here is a math link that starts off with the basic equations.
     http://www.sonoma.edu/users/w/wilsonst/papers/geometry/circles/default.html

BTW - some notation for writing on EE to show raising to the power of 2:
(x-2)^2 +(y+2)^2 = 1
x^2 +y^2 = 4

You can even do:
(x-2)² +(y+2)² = 1
x² +y² = 4
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tango2009Author Commented:
This is how far Iv got can you see if I am doing this right

(x-2) ^2 + (y+2)^2 = 1
x^2 + y^2 = 4

(x + 2 + y + 2 -1) + (x^2 + y^2 - 4)

x + y + 4x + 4y

x + y -8

= 4 (x + y -2 )

Y = (-4x + 8) / 4 = -x + 2 -> 2 - x

next I have to substitute the value of y in the eqaution. I am on the right tracks so far or am I off point.
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ozoCommented:
What do those lines represent?
What's on the left of the 3rd = ?
Do you need to find the points of intersection, or just determine whether they intersect?
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tango2009Author Commented:
These are two circles that I have to check if they intersect or not?

'What's on the left of the 3rd = ?' Not sure which one you mean. Is it this bit =4 ( x + y - 2) if so thats a typo sorry there shouldn't be an eqauls there.
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d-glitchConnect With a Mentor Commented:
You can tell (just by looking at the two equations) that the radius of the first circle is 1
and the radius of the second circle is 2.  Do you see this?

You can also tell (again just by looking) what the centers of the two circles are.
Do you know how to do this?
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phoffricCommented:
In my previous link is this link:
     http://www.sonoma.edu/users/w/wilsonst/papers/geometry/circles/T1--2/T1-3-1.html
which helps you understand the coordinates of the center of a circle and the radius.
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tango2009Author Commented:
Ok

So with this eqaution (x-2)² +(y+2)² = 1

The centre of the circle is (2,2) and the radius is 1

With the other equation isn't the radius 4 and the centre (2,2)?
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phoffricCommented:
>> the radius is 1
Yes

>> The centre of the circle is (2,2)
No. Take a look at the formulas and be careful about the signs.

Hint:  You probably know this:    +9 = -(-9)
But see if you can apply this idea to the center of circle problem.
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phoffricCommented:
>> x^2 + y^2 = 4
>> isn't the radius 4 and the centre (2,2)?

Ok, you are saying that r = 4, x0 = y0 = 2.
Test out your idea by plugging in your values into the circle equation from the link:

(x - x0)² + (y - y0)² = r²

Write down the equation plugging in the values. The equation you come up with will be a circle whose center is (2,2) and whose radius is 4. Does this equation match the equation of the one you posted?
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phoffricConnect With a Mentor Commented:
Here is an example that I am making up.(x + 19)² + (y - 17)² = 36We recognize the form of this equation as a circle having a center (x0,y0) and radius r. We want to find out these 3 values.The standard form of a circle (from the link) is:    (x - x0)²  + (y - y0)² =   r²    (x + 19)² + (y - 17)² = 36We need minus signs inside the parenthesis. Notice that +19 == -(-19), and that 36 == 6² , so rewrite as:    (x + 19)² + (y - 17)² = 36   ==>    (x - (-19) )²  + (y - 17)² =   6²    (x -   x0  )²  + (y - y0)² =   r²By inspection, you can see that x0 is -19, y0 is 17, and r is 6; so we have a circle whose center is    (-19, 6) and whose radius is 6.
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phoffricCommented:
correction to cut and paste error:By inspection, you can see that x0 is -19, y0 is 17, and r is 6; so we have a circle whose center is   (-19, 17) and whose radius is 6.
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nickalhConnect With a Mentor Commented:
To extract the center and radius from a circle convert it to

GENERIC EQUATION OF A CIRCLE
(x - x_0)² + (y - y_0)² = r²
form

For example,
(x - 3)^2  + (y+ 7)^2 = 100
becomes

(x - 3)^2  + (y  - (- 7) )^2 = 10^2
Now it's identically in the generic form.
We can lift out x_0 = 3,   y_0= -7 and the radius = 10.


So for the 2nd equation, x^2 + y^2 = 4
the center is not (2, 2) because there is no 2 being added or subtracted from x and no 2 being added or subtracted from y.

The radius is not 4.  because it still needs to go into square form first.  Only when we have the form
(x - x_0)² + (y - y_0)² = r²
can we directly find x_0, y_0 or r from the equation.
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