What is theslope of the line joining the points of intersection of the graphs x^2+y^2=4 and (x-2)^2+y^2=4?

Answer 1

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

First, let's find the points of intersection. We can subtract the first equation from the second to eliminate the y variable.

x^2 - (x - 2)^2 = 0

Factor as a difference of squares,

(x - (x - 2)) (x + (x - 2)) = 0

(x - x + 2)(2x - 2) = 0
2(2x - 2) = 0
2(2)(x - 1) = 0
x - 1 = 0
x = 1

Plugging x = 1 into the first equation, x^2 + y^2 = 4, we get

1^2 + y^2 = 4
1 + y^2 = 4
y^2 = 3
y = +/- sqrt(3)

Therefore, the two points of intersection are (1, sqrt(3)) and (1, -sqrt(3))

We can already tell by inspection that they lie on the same vertical line, meaning the slope is going to be undefined.

Answer 2

x^2+y^2=4 and x^2 -4x +4+y^2=4
Eqt 1 => x^2 -4x +4+y^2=4
Eqt 2 => x^2+y^2=4.............subtract 1 from 2
.........=> -4x +4 = 0
=> x = 1 this line is // to Y axis
is the line joining the points of intersection
=> slope = oo (undefined)

Answer 3

These two circles have the same y-coordinate of the center. By symmetry, the two intersection points have the same x value.Therefore, the slope of the line joining the points of intersection is undefined or infinitely large.

Answer 4

x^2+y^2=4 and (x-2)^2+y^2=4
(x-2)^+4-x^2 = 4
x^2-4x+4+4-x^2 = 4
-4x = -4 x = 1
When x= 1, y= +/- sqrt(3)
The points of intersection are thus (1,sqrt(3)) and (1-sqrt(3)).
The slope of the line joining these two points is
[-sqrt(3) -sqrt(3)]/(1-) = undefined.

Answer 5

The points of intersection are (1, √3) and (1, -√3)

The line of intersection is x = 1

The gradient of this line is ∞