Đường tròn (C): x^2 +y^2 +12x-14y+4=0 có dạng tổng quát là

A. $\left(C\right): {\left(x+6\right)}^2+{\left(y-7\right)}^2=9$

B. $\left(C\right): {\left(x+6\right)}^2+{\left(y-7\right)}^2=81$

C. $\left(C\right): {\left(x+6\right)}^2+{\left(y-7\right)}^2=89$

D. $\left(C\right): {\left(x+6\right)}^2+{\left(y-7\right)}^2=\sqrt{89}$