Question

I need help with this question (with the questions under the equation)

Answer 1

Option C

$(x-2)^2+(y-2)^2=8_{}$$\begin{gathered} r^2=(0-2)^2+(4-2)^2 \\ r^2=(-2)^2+(-2)^2 \\ r^2=4+4\text{ = 8} \end{gathered}$$\begin{gathered} \text{Equation of a circle = (x-a)}^2+(y-b)^2=r^2 \\ a=\text{ }\frac{0+4}{2}=\frac{4}{2}=2 \\ b=\frac{4+0}{2}=\frac{4}{2}=2 \end{gathered}$$\text{Equation of a circle = (x-2)}^2+^2(y-2)^2=8$