## College Algebra (10th Edition)

$(x-2)^2+(y-2)^2=5$
The midpoint of the diameter's endpoints is the circle's center. RECALL: The coordinates of the midpoint of the segment connecting the points $(x_1, y_1)$ and $(x_2,y_2)$ is given by the formula: $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ Find the center of the circle by solving for the coordinates of the midpoint of the diameter using the midpoint formula above to obtain: $\text{center} = \left(\frac{4+0}{2}, \frac{3+1}{2}\right)=\left(\frac{4}{2}, \frac{4}{2}\right)=(2, 2)$ With its center at $(2, 2)$, the tentative equation of the circle is: $(x-h)^2+(y-k)^2=r^2 \\(x-2)^2+(y-2)^2=r^2$ Find the value of $r^2$ by substituting the x and y values of a point on the circle. Since $(0, 1)$ is an endpoint of the circle's diameter, this point is on the circle. Substitute the x and y values of this point into the tentative equation above to obtain: $(x-2)^2+(y-2)^2=r^2 \\(0-2)^2+ (1-2)^2=r^2 \\(-2)^2 + (-1)^2 = r^2 \\4+1=r^2 \\5=r^2$ Therefore, the equation of the circle is: $(x-2)^2+(y-2)^2=5$