What is the distance between the center of the circle with equation $x^2+y^2=2x+4y-1$ and the point $(13,7)$?
I notice that the equation of the circle is not in the standard form $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. To get it into that form, I need to complete the square for both $x$ and $y$. I rewrite the equation as $(x^2-2x)+(y^2-4y)=-1$ and then add the appropriate constants to both sides to make perfect squares. For $x$, I need to add $(2/2)^2=1$ to both sides, and for $y$, I need to add $(4/2)^2=4$ to both sides. This gives me $(x^2-2x+1)+(y^2-4y+4)=-1+1+4$. Simplifying, I get $(x-1)^2+(y-2)^2=4$. Now I can see that the center of the circle is $(1,2)$ and the radius is $2$. To find the distance between the center and the point $(13,7)$, I can use the distance formula, which is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Plugging in the coordinates, I get $d=\sqrt{(13-1)^2+(7-2)^2}$. Simplifying, I get $d=\sqrt{144+25}$. Taking the square root, I get $d=\sqrt{169}$. Finally, I get $d=13$. 