How do you solve #x^2 + y^2 = 9# and #x^2 – 3y = 9#?

1 Answer
Sep 4, 2015

Answer:

Solution is (-3,0), (3,0) and (0,-3)

Explanation:

#x^2+y^2 =9# represents a circle centered at (0,0) with radius 3 and #x^2-3y=9# represents a vertical parabola opening up with its vertex at (0,-3). These intersect at 3 points which can be determined graphically or algebraically.

Solving algebraically, it would be #3y+9+y^2=9#, which means y(y+3)=0. Thus y=0, -3

For y=0, the equation of circle gives x= 3, -3, hence two of the points are (-3,0) and (3,0)

For y=-3, the circle equation gives x=0, hence the third point is (0,-3)