Answer :
Find the
center and
Radius (1,2) (1,0) write the standard form of the equation
Step 1
by the graph, let's assume the coordinates are of the center and an external point of the circle.
find the radius, use the distance between two points formula
remember
[tex]\begin{gathered} \text{let} \\ P1(x_1,y_1)andP2(x_2,y_2) \end{gathered}[/tex]the distance between P1 and P2 is given by:
[tex]d=\sqrt{(y_2-y_1_{})^2+(x_2-x_1_{}_{})^2}[/tex]replace the values
let P1(1,2) and P2(1,0)
[tex]\begin{gathered} d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2} \\ d=\sqrt{(0-2)^2+(1-1)^2} \\ d=\sqrt{4} \\ d=2 \end{gathered}[/tex]therefore, the radius is 2
Step 2
the standard form of the equation of a circle is
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ \text{where (h,k) is the center of the circle and r is the radius} \end{gathered}[/tex]by the graph, the center is (1,2), so replace
[tex]\begin{gathered} (x-1)^2+(y-2)^2=2^2 \\ (x-1)^2+(y-2)^2=4 \end{gathered}[/tex]and that's all, it would be your answer. I hope it helps you