intersect x^2+y^2=36-8x^2-8y^2
with .... ?
x^2+y^2=36-8x^2-8y^2
x^2+8x^2+y^2+8y^2=36
9x^2+9y^2=36 Divide both sides with 9
x^2+y^2=4
That is circle.
x^2+y^2=R^2
Coordinates of center: 0,0
Radius: R=2
Intersect:
For x=0 y^2=4, y=sqrt(4)= +or- 2
For y=0 x^2=4, x=sqrt(4)= +or- 2
To find the intersection points of the given equation, we need to solve for the value(s) of x and y that satisfy the equation:
x^2 + y^2 = 36 - 8x^2 - 8y^2
First, let's simplify the equation by combining like terms:
9x^2 + 9y^2 = 36
Next, divide both sides of the equation by 9 to isolate the variables:
x^2 + y^2 = 4
Now, let's rearrange the equation so that we have y in terms of x:
y^2 = 4 - x^2
Taking the square root of both sides gives:
y = ± √(4 - x^2)
Now, we have two separate equations to find the possible values of y:
1. y = √(4 - x^2)
2. y = -√(4 - x^2)
To visualize the solution, we can graph the equations:
- The equation y = √(4 - x^2) represents the upper half of a circle centered at (0, 0) with a radius of 2.
- The equation y = -√(4 - x^2) represents the lower half of the same circle.
The intersection points occur where the circle intersects the x-axis.
To find the x-coordinate of the intersection points, we can set the y-value to 0 in each equation:
1. y = √(4 - x^2): Set y = 0
0 = √(4 - x^2)
Squaring both sides of the equation gives:
0 = 4 - x^2
Rearranging the equation and solving for x:
x^2 = 4
Taking the square root of both sides:
x = ± 2
Therefore, the x-coordinates of the intersection points are x = 2 and x = -2.
To find the y-coordinate of each intersection point, substitute the x-coordinates into either of the original equations:
For x = 2:
y = √(4 - 2^2)
= √(4 - 4)
= √0
= 0
So, one intersection point is (2, 0).
For x = -2:
y = √(4 - (-2)^2)
= √(4 - 4)
= √0
= 0
So, the other intersection point is (-2, 0).
Therefore, the equation x^2 + y^2 = 36 - 8x^2 - 8y^2 intersects the x-axis at the points (2, 0) and (-2, 0).