circle: x^2 + y^2 = 9
parabola: y = x^2
using substitution
y + y^2 = 9
y^2 + y - 9 = 0
this quadratic will have 2 real solutions,
so there must be 2 intersection points
(make a quick sketch to verify this for yourself )
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?
2 answers
two solutions is correct, but the equation for a circle is:
(x-a)^2 + (y-b)^2 = r^2; where (a, b) is the center of the circle, so
(x-0)^2 + (y-1)^2 = 9
The points of intersection are:
(1.836377228, 3.372281323) and
(-1.836377228, -1.372281323)
(x-a)^2 + (y-b)^2 = r^2; where (a, b) is the center of the circle, so
(x-0)^2 + (y-1)^2 = 9
The points of intersection are:
(1.836377228, 3.372281323) and
(-1.836377228, -1.372281323)
1 answer