Answer
center: $(2.5, 2)$;
radius = $1.5$
standard form of the equation: $\color{blue}{(x-2.5)^2+(y-2)^2=2.25}$
Work Step by Step
RECALL:
The standard form of a circle's equation is:
$(x-h)^2 +(y-k)^2=r^2$
where $r$ = radius and $(h, k)$ is the center.
The points on the circle $(1, 2)$ and $(4, 2)$ belong to the same horizontal line.
This means that the segment connecting them is a diameter of the circle. The center of the circle is the midpoint of the diameter.
Find the midpoint using the midpoint formula $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$:
center = $\left(\dfrac{1+4}{2}, \dfrac{2+2}{2}\right)=(2.5, 2)$
Substitute $2.5$ to $h$ and $2$ to $k$ in the standard form above to obtain the tentative equation:
$(x-2.5)^2+(y-2)^2=r^2$
To find the value of $r$, substitute the x and y values of the point $(1, 2)$ into the tentative equation above to obtain:
$(x-2.5)^2+(y-2)^2=r^2
\\(1-2.5)^2+(2-2)^2=r^2
\\(-1.5)^2+0^2=r^2
\\2.25 + 0 = r^2
\\2.25 = r^2
\\\pm\sqrt{2,25} = \sqrt{r^2}
\\\pm 1.5 = r$
Since the radius cannot be negative, $r=1.5$
Therefore, the standard form of the circle's equation is:
$(x-2.5)^2+(y-2)^2=(1.5)^2
\\\color{blue}{(x-2.5)^2+(y-2)^2=2.25}$
