jmc

Add $(-2/2{)}^2$ and $(-10/2{)}^2$ to the first equation and $(-8/2{)}^2$ and $(-10/2{)}^2$ to the second equation to find that the given equations are equivalent to $$\begin{array}{rl}(x^2-2x+1)+(y^2-10y+25)& =1\text{, and}\\ (x^2-8x+16)+(y^2-10y+25)& =4\end{array}$$ which are equivalent to $$\begin{array}{rl}(x-1{)}^2+(y-5{)}^2& =1^2,\\ (x-4{)}^2+(y-5{)}^2& =2^2,\end{array}$$ respectively. Hence, the two circles have centers $(1,5)$ and $(4,5)$ respectively and radii $1$ and $2$ respectively. Since the centers of the circles are $3$ units apart and the sum of their radii is $3$, the two circles intersect at only one point. We can see that $(2,5)$ is the desired intersection point, so our product is $2\cdot 5=\boxed{10}$.