A circle in the xy -plane has a diameter with endpoints 2 , 4 and 2

The correct answer is $5$. The standard form of an equation of a circle in the xy-plane is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where $h$, $k$, and $r$ are constants, the coordinates of the center of the circle are $\left(h,k\right)$, and the length of the radius of the circle is $r$. It′s given that an equation of the circle is ${\left(x-2\right)}^2+{\left(y-9\right)}^2=r^2$. Therefore, the center of this circle is $\left(2,9\right)$. It’s given that the endpoints of a diameter of the circle are $\left(2,4\right)$ and $\left(2,14\right)$. The length of the radius is the distance from the center of the circle to an endpoint of a diameter of the circle, which can be found using the distance formula, $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$. Substituting the center of the circle $\left(2,9\right)$ and one endpoint of the diameter $\left(2,4\right)$ in this formula gives a distance of $\sqrt{{\left(2-2\right)}^2+{\left(9-4\right)}^2}$, or $\sqrt{0^2+5^2}$, which is equivalent to $5$. Since the distance from the center of the circle to an endpoint of a diameter is $5$, the value of $r$ is $5$.