n the xy -plane, the graph of y = x^2
and the circle with
center (0,1) and radius 3 have how many points of intersection?
To find the number of points of intersection between the graph of the equation y = x^2 and the circle with center (0,1) and radius 3 in the xy-plane, we need to solve the system of equations.
The equation of the circle with center (0, 1) and radius 3 can be written as:
(x - 0)^2 + (y - 1)^2 = 3^2
Simplifying, we get:
x^2 + (y - 1)^2 = 9
The equation of the graph of y = x^2 is already given.
To solve this system of equations, we can substitute the value of y in the equation of the circle from the equation of the graph:
x^2 + (x^2 - 1)^2 = 9
Now, we can solve this equation algebraically or using a graphing calculator to find the intersections.
Algebraically:
x^4 - 2x^2 + 1 + x^2 = 9
x^4 - x^2 - 8 = 0
This equation can be factored as:
(x^2 - 4)(x^2 + 2) = 0
Solving for x, we have two cases:
Case 1: x^2 - 4 = 0
x^2 = 4
x = ±2
Case 2: x^2 + 2 = 0 (no real solutions)
Since x^2 + 2 = 0 has no real solutions, we only consider the solutions from Case 1.
For x = 2, substituting back into the equation of the circle:
2^2 + (y - 1)^2 = 9
4 + (y - 1)^2 = 9
(y - 1)^2 = 5
y - 1 = ±√5
y = 1 ± √5
For x = -2, substituting back into the equation of the circle:
(-2)^2 + (y - 1)^2 = 9
4 + (y - 1)^2 = 9
(y - 1)^2 = 5
y - 1 = ±√5
y = 1 ± √5
So, we have four points of intersection: (2, 1 + √5), (2, 1 - √5), (-2, 1 + √5), and (-2, 1 - √5).