Question

The equation of circle having a diameter with endpoints (-3, 1) and (5, 7) is

Answer 1

The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the center
r - the radius

The center is the midpoint of the diameter.
$(-3,1) \\ x_1=-3 \\ y_1=1 \\ \\ (5,7) \\ x_2=5 \\ y_2=7 \\ \\ \hbox{the midpoint: } (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ \frac{x_1+x_2}{2}=\frac{-3+5}{2}=\frac{2}{2}=1 \\ \frac{y_1_y_2}{2}=\frac{1+7}{2}=\frac{8}{2}=4 \\ \Downarrow \\ (1,4) \\ h=1 \\ k=4$

The radius is half the diameter, or the distance from one of the endpoints to the center.
$r=\sqrt{(x_1-h)^2+(y_1-k)^2}=\sqrt{(-3-1)^2+(1-4)^2}=\\ =\sqrt{(-4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5 \\ \Downarrow \\ r^2=5^2=25$

The equation of the circle:
$\boxed{(x-1)^2+(y-4)^2=25}$

Answer 2

Answer: B.) (x - 1)² + (y - 4)² = 25