Question 1011104
Find the center({{{h}}},{{{k}}}) and {{{r}}} of the circle

{{{x^2+y^2+10x-18y+106=49}}}

{{{(x^2+10x)+(y^2-18y)+106=49}}}......complete squares

{{{(x^2+10x+b^2)-b^2+(y^2-18y+b^2)-b^2+106=49}}}

{{{(x^2+10x+5^2)-5^2+(y^2-18y+9^2)-9^2+106=49}}}

{{{(x+5)^2-25+(y-9)^2-81+106=49}}}

{{{(x+5)^2+(y-9)^2-cross(106)+cross(106)=49}}}

{{{(x+5)^2+(y-9)^2=7^2}}}

so,{{{h=-5}}},{{{k=9}}}, {{{r=7}}}, and the center is at ({{{-5}}},{{{9}}})


{{{drawing( 600, 600, -20, 10, -10, 20,
circle(-5,9,.12),locate(-5,9,C(-5,9)),circle(-5,9,7),
 graph( 600, 600, -20, 10, -10, 20,-sqrt(-(x+5)^2+49)+9, sqrt(-(x+5)^2+49)+9)) }}}