In the xy-plane, the graph of y = x² and the circle with center (0, 1) and the radius 3 have how many points of intersection?

Solution:

The equation of the circle can be written as follows:

If a point P(x, y) exists on the circle, then

(x - 0)² + (y - 1)² = (3)² [(x - h)² + (y - k)² = r; (h, k) = coordinates of center and r = radius]

x² + y² - 2y + 1 = 9

x² + y² - 2y = 8

Since y = x², substituting in the above expression

y + y² - 2y = 8

y² - y = 8

y² - y - 8 = 0

Since the above equation is a quadratic equation it will give two values of y and hence there will be two corresponding values of x.

Hence we can conclude that the circle with center (0,1) and radius 3 will intersect the curve at two points only.

In the xy-plane, the graph of y = x² and the circle with center (0, 1) and the radius 3 have how many points of intersection

Summary:

In the xy-plane, the graph of y = x² and the circle with center (0, 1) and the radius 3 have two points of intersection.