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

Guest Sep 2, 2021

#1**+1 **

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.

**y**^{2} - **6y** + **9** = **9**

**y**^{2} + **8y** + **16** = **16**

We can also eliminate the term **y**^{2} 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)

helperid1839321 Sep 2, 2021