Find the intersection points of the circles (x - 2)^2 + (y - 3)^2 = 9 and (x - 2)^2 + (y + 4)^2 = 16.
+633 This one appears solvable without graphing or using a calculator due to one quirk:
Take another look at those equations for the circles
(x-2)2 + (y-3)2 = 9
(x-2)2 + (y+4)2 = 16
The x-term in each equation is actually identical. We can therefore eliminate each from the equation for the purposes of solving for y.
We'll also expand the y-term here for convenience in the next step.
y2 - 6y + 9 = 9
y2 + 8y + 16 = 16
We can also eliminate the term y2 here as it's shared on both sides. We can also take out the constant values on both sides, giving us...
-6y = 0
8y = 0
Through multiplication of each equation we can get y = 0. Now plugging this into the original equations...
(x-2)2 + (0-3)2 = 9
(x-2)2 + (0+4)2 = 16
(x-2)2 + (-3)2 = 90
(x-2)2 + (4)2 = 16
(x-2)2 + 9 = 9
(x-2)2 + 16 = 16
(x-2)2 = 0
(x-2)2 = 0
Taking square root of both sides...
x-2 = 0
x-2 = 0
x = 2
x = 2
Going back, we now have 2 solutions for (2, 0). Thus, the question is worded misleadingly, and the only intersection is at (2, 0)
