how do you find three solutions algebraically for x^2+y^2=16 and y=x^2-4

Question

intersection of parabola and circle,solve showing all work

Answer 1

x²+y²=16 is the circle, radius 4, with centre at the origin (0,0). y=x²-4 is the parabola, vertex at (0,-4), which happens to lie on the circumference of the circle.

x²+x⁴-8x²+16=16, x⁴-7x²=0=x²(x²-7)=x²(x-√7)(x+√7), so x=0 (y=-4), √7 (y=3) -7 (y=3).

The intersections are (0,-4), (√7,3), (-√7,3).