How do you solve #x^2 + y^2 = 9# and #x^2 – 3y = 9#?
1 Answer
Sep 4, 2015
Solution is (-3,0), (3,0) and (0,-3)
Explanation:
Solving algebraically, it would be
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)