Answer
The center of the circle is (1, 3) and its radius is 6
Step-by-step explanation
Given the equation:
[tex]x^2-2x+y^2-6y=26[/tex]we need to complete the square of the following expressions:
[tex]\begin{gathered} x^2-2x \\ y^2-6y \end{gathered}[/tex]To do this we need to divide the coefficient of the x-term, which is -2, and the y-term, which is -6, divide them by 2, and then square them, that is,
[tex]\begin{gathered} (-\frac{2}{2})^2=(-1)^2=1 \\ (-\frac{6}{2})^2=(-3)^2=9 \end{gathered}[/tex]Adding 1 and 9 at both sides of the equation:
[tex]\begin{gathered} x^2-2x+y^2-6y+1+9=26+1+9 \\ (x^2-2x+1)+(y^2-6y+9)=36 \\ (x-1)^2+(y-3)^2=36 \end{gathered}[/tex]This equation has the form:
[tex](x-h)^2+(y-k)^2=r^2[/tex]which is a circle centered at (h, k) with radius r.
Therefore, the center of the circle is (1, 3) and it radius is 6 (6² = 36)
