What is the product of all the coordinates of all the points of intersection of the two circles defined by $x^2-2x +y^2-10y+25=0$ and $x^2-8x+y^2-10y+37=0$?
I notice that both circles have the same $y$-term, $y^2-10y$, so I can try to eliminate that by subtracting one equation from the other. Doing that, I get $6x-12=0$, which implies that $x=2$. That means that all the points of intersection have $x=2$, so I can plug that into either equation to find the possible values of $y$. Using the first equation, I get $4-4+y^2-10y+25=0$, which simplifies to $y^2-10y+25=0$. This is a quadratic equation that I can solve by factoring. I look for two numbers that multiply to $25$ and add to $-10$, and I find that $-5$ and $-5$ work. So $y^2-10y+25=(y-5)^2=0$, which implies that $y=5$. Therefore, there is only one point of intersection, and it is $(2,5)$. The product of the coordinates of this point is $2 \cdot 5 = 10$. 