$\text{a)}\\\text{Tâm đường tròn: }\begin{cases}x_I=\dfrac{-1+5}{2}=2\\y_I=\dfrac{1+3}{2}=2\end{cases}\\\to I(2; 2)\\\overrightarrow{AB}=(6; 2) \to AB=\sqrt{6^2+2^2}=2\sqrt{10}\to R=\dfrac{1}{2}AB=\sqrt{10}\\\to \text{Phương trình (C): }(x-2)^2+(y-2)^2=10 \\\to x^2+y^2-4x-4y-2=0\\\text{b)}\\\text{Vì Δ tiếp xúc đường tròn }\\\to d(I; \Delta)=\dfrac{\Big|-1+2.(-2)+7\Big|}{\sqrt{1^2+(-2)^2}}=\dfrac{2\sqrt{5}}{5}=R\\\to \text{Phương trình (C): }(x+1)^2+(y-2)^2=\Bigg(\dfrac{2\sqrt{5}}{5}\Bigg)^2\\\to x^2+y^2+2x-4y+\dfrac{21}{5}=0$