Question

Find the intersection points, x^2+y^2-16y+39=0, y^2-x^2-9=0

Answer 1

X^2 + y^2 - 16y + 39 = 0 
-x^2 + y^2 - 9 = 0 
----------------------------------- 
2y^2 - 16y + 30 = 0 iff 
y^2 - 8y + 15 = 0 iff 
(y - 3)(y - 5) = 0 iff 
y = 3 or y = 5. 

Now plug in y value into any of the given equations (the second one is easier) to find respective x values. 

y^2 - x^2 - 9 = 0 
(3)^2 - x^2 - 9 = 0 iff 
9 - x^2 - 9 = 0 iff 
x^2 = 0 iff 
x = 0 
(0, 3) 

y^2 - x^2 - 9 = 0 
(5)^2 - x^2 - 9 = 0 iff 
25 - x^2 - 9 = 0 iff 
16 - x^2 = 0 iff 
(4 - x)(4 + x) = 0 iff 
x = 4 or x = -4 
(4, 5), (-4, 5) 

We have three points of intersection, which are (0, 3), (4, 5), and (-4, 5).