We have to find the radius and center of the circle from the given equation.
The simplest way to do it is to rearrange the equation in the form:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
which correspond to a circle with center (a,b) and radius r.
Then, we start by writing the equation:
[tex]x^2-2x+y^2-6y=26[/tex]
We have to complete the squares for both x and y, so we can do it as:
[tex]\begin{gathered} (x^2-2\cdot1\cdot x+1^2-1^2)+(y^2-2\cdot3\cdot y+3^2-3^2)=26 \\ (x-1)^2-1+(y-3)^2-9=26 \\ (x-1)^2+(y-3)^2=26+1+9 \\ (x-1)^2+(y-3)^2=36 \\ (x-1)^2+(y-3)^2=9^2 \end{gathered}[/tex]
We have arrived to the equation form we needed. In this equation we can see that the center is (1,3) and the radius is r = 9.
Answer: Center is (1,3) and radius is 9.
