Answer:
Center (1,3) and radius 6
Step-by-step explanation:
We must complete the square to find the center and radius of the circle.
First make sure the x and y squared terms have 1 as their coefficients. We also make sure x and y terms together.
[tex]x^2-2x+y^2-6y=26[/tex]
We now create space between the x and y terms with parenthesis.
[tex](x^2-2x)+(y^2-6y)=26[/tex]
We complete the square by taking the middle terms -2x and the -6y - divide each and square them.
[tex]\frac{-2}{2} =(-1)^{2} =1[/tex]
[tex]\frac{-6}{2} =(-3)^{2} =9[/tex]
We add the squares to both sides.
[tex](x^2-2x+1)+(y^2-6y+9)=26+1+9[/tex]
Simplify.
[tex](x^2-2x+1)+(y^2-6y+9)=36[/tex]
And write the quadratics in factored form.
[tex](x-1)^{2} +(y-3)^{2} =36[/tex]
The center is (h,k) or (1,3). The radius is the square root of 36 which is 6.
