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