Find the points of intersections of the circles:x^2 + y^2 - 2x -2y -2 = 0 and x^2 + y^2 + 2x + 2y -2 = 0. Draw the circles.
intersections:
x^2 + y^2 - 2x - 2y -2 = 0
x^2 + y^2 + 2x + 2y -2 = 0
---------------------------subtract
-4x -4y = 0
x = - y so along that line of slope -1
now where does x = -y on either circle?
2 y^2 + 2y -2y = 2
y^2 = 1
y = +/- 1
x = -/+ 1 in other words at (-1,1) and (1,-1)
to draw them:
x^2 - 2 x = -y^2+2y +2
x^2 - 2 x + 1 = -y^2+ 2 y + 3
(x-1)^2 = -y^2+2y+3
y^2 - 2 y = 3 - (x-1)^2
y^2 - 2 y + 1 = 4 -(x-1)^2
so when all is said and done
(x-1)^2 + (y-1)^2 = 2^2
circle of radius 2 with center at (1,1)
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x^2 + 2x = -y^2 -2y + 2
x^2 + 2 x + 1 = -y^2 - 2 y + 3
-(x+1)^2 +3 = y^2 + 2y
-(x+1)^2 +4 = (y+1)^2
again radius = 2 but center at (-1,-1)
To find the points of intersection between the two circles, we need to solve the system of equations representing the two circle equations:
1) x^2 + y^2 - 2x - 2y - 2 = 0
2) x^2 + y^2 + 2x + 2y - 2 = 0
First, let's rearrange equation 1) and equation 2) to get the equations in standard form:
1) x^2 - 2x + y^2 - 2y = 2
2) x^2 + 2x + y^2 + 2y = 2
The standard form of a circle equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
Comparing equations 1) and 2) with the standard form, we can see that:
1) (x - 1)^2 + (y - 1)^2 = 4
2) (x + 1)^2 + (y + 1)^2 = 4
So, both circles have a radius of 2 and their centers are (1, 1) and (-1, -1) respectively.
To find the points of intersection, we can set the two equations equal to each other:
(x - 1)^2 + (y - 1)^2 = (x + 1)^2 + (y + 1)^2
Expanding both sides:
x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 + 2x + 1 + y^2 + 2y + 1
Simplifying:
-4x - 4y + 2 = 4x + 4y + 2
Rearranging terms:
-8x - 8y = 0
Dividing both sides by -8:
x + y = 0
So, the points of intersection lie on the line x + y = 0.
Now, let's draw the circles:
The circle with center (1, 1) and radius 2 is a circle centered at (1, 1) with a radius of length 2. Similarly, the circle with center (-1, -1) and radius 2 is a circle centered at (-1, -1) with a radius of length 2.