# 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?
d-glitch

That's only one circle.
The way to check if two circles intersect is to look at the distance between their centers
and the sum of their radii.
Do you know how to find the center and radius of this circle:

(x-2)² +(y+2)² = 1
phoffric

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.

sorry made a mistake in posting its these two circles

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

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

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.
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?

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.
ozo

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SOLUTION

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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.

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)?
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.
>> 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.

(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?
SOLUTION

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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.
SOLUTION

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