What are the points of intersection of the equation of the circles x2 + 8x + y2 - 16y = 0 and x2 + 8x + y2 - 8y = 32

We have that
x² + 8x + y² - 16y = 0
Group terms that contain the same variable
(x²+8x)+(y²-16y)=0
Complete the square twice. Remember to balance the equation by adding the same constants to each side
(x²+8x+16)+(y²-16y+64)=16+64

Rewrite as perfect squares

(x+4)²+(y-8)²=80
circle with a center (-4,8) and radius √80 units

and

x² + 8x + y² - 8y = 32
Group terms that contain the same variable
(x² + 8x) + (y² - 8y) = 32
Complete the square twice. Remember to balance the equation by adding the same constants to each side
(x² + 8x+16) + (y² - 8y+16) = 32+16+16

Rewrite as perfect squares

(x+4)² + (y-4)² =64
circle with a center (-4,4) and radius 8 units

using a graph tool
see the attached figure

the solution are the points
(-12,4) and (4,4)

Bis the right answer

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