## College Algebra (10th Edition)

$(x+1)^2+(y−3)^2=5$
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: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ Find the center of the circle by solving for the coordinates of the midpoint of the diameter using the midpoint formula above to obtain: center=$(\frac{1+(−3)}{2},\frac{4+2}{2})=(−\frac{2}{2},\frac{6}{2})=(−1,3)$ With its center at $(−1,3)$, the tentative equation of the circle is: $(x−h)^2+(y−k)^2=r^2 \\$ $[x−(−1)]^2+(y−3)^2=r^2 \\(x+1)^2+(y−3)^2=r^2$ Find the value of $r^2$ by substituting the x and y values of a point on the circle. Since $(1,4)$ 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: $(x+1)^2+(y−3)^2=r^2 \\(1+1)^2+(4−3)^2=r^2 \\2^2+1^2=r^2 \\4+1=r^2 \\5=r^2$ Therefore, the equation of the circle is: $(x+1)^2+(y−3)^2=5$