center (1,3) with 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.
![x^2-2x+y^2-6y=26](/tex.php?f=x^2-2x+y^2-6y=26)
we now create space between the x and y terms with parenthesis.
![(x^2-2x)+(y^2-6y)=26](/tex.php?f=(x^2-2x)+(y^2-6y)=26)
we complete the square by taking the middle terms -2x and the -6y - divide each and square them.
![\frac{-2}{2} =(-1)^{2} =1](/tex.php?f=\frac{-2}{2} =(-1)^{2} =1)
![\frac{-6}{2} =(-3)^{2} =9](/tex.php?f=\frac{-6}{2} =(-3)^{2} =9)
we add the squares to both sides.
![(x^2-2x+1)+(y^2-6y+9)=26+1+9](/tex.php?f=(x^2-2x+1)+(y^2-6y+9)=26+1+9)
simplify.
![(x^2-2x+1)+(y^2-6y+9)=36](/tex.php?f=(x^2-2x+1)+(y^2-6y+9)=36)
and write the quadratics in factored form.
![(x-1)^{2} +(y-3)^{2} =36](/tex.php?f=(x-1)^{2} +(y-3)^{2} =36)
the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.