Answer
$(x+1)^2+(y−3)^2=5$
Work Step by Step
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:
$(x+1)^2+(y−3)^2=5$
