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# Coordinates

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Find the intersection points of the circles (x - 2)^2 + (y - 3)^2 = 9 and (x - 2)^2 + (y + 4)^2 = 16.

Sep 2, 2021

#1
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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

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)

Sep 2, 2021