What is the center of a circle whose equation is x^2+y^2-12x-2y+12=0

Answer:[tex](6,\, 1)[/tex].Step-by-step explanation:Let [tex](a,\, b)[/tex] denote the center of this circle (where [tex]a [/tex] and [tex]b[/tex] are to be found.) If the radius of this circle is [tex]r[/tex], the equation of this circle would be:[tex](x - a)^{2} + (y - b)^{2} = r^{2}[/tex].Expand this equation using binomial theorem:[tex]x^{2} - 2\, a\, x + a^{2} + y^{2} - 2\, b\, y + b^{2} = r^{2}[/tex].Both this equation and the given equation [tex]x^{2} + y^{2} - 12\, x - 2\, y + 12 = 0[/tex]  describe the same circle. Therefore, corresponding coefficients of the two equations should match one another:The coefficient of the [tex]x[/tex] term should match in the two equations. Therefore: [tex](-2\, a) = (-12)[/tex].The coefficient of the [tex]y[/tex] term should also match in the two equations. Therefore: [tex](-2\, b) = (-2)[/tex].Solve these two equation for [tex]a[/tex] and [tex]b[/tex]: [tex]a = 6[/tex] and [tex]b = 1[/tex].Substitute [tex]a = 6[/tex] and [tex]b = 1[/tex] into [tex](a,\, b)[/tex] to find the center of this circle: [tex](6,\, 1)[/tex].