Question

Find an equation of the circle that satisfies the given conditions.

Answer 1

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

The midpoint of the diameter is the centre of a circle.
The coordinates of the midpoint:
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
(x₁,y₁), (x₂,y₂) - the coordinates of endpoints

$P(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ Q(5,9) \\ x_2=5 \\ y_2=9 \\ \\ \frac{x_1+x_2}{2}=\frac{-1+5}{2}=\frac{4}{2}=2 \\ \frac{y_1+y_2}{2}=\frac{1+9}{2}=\frac{10}{2}=5$

The centre of the circle is (2,5).

The radius is the distance between an endpoint of the diameter and the centre.
The formula for distance:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(-1,1) \\ x_1=-1 \\ y_1=1 \\ \\ (2,5) \\ x_2=2 \\ y_2=5 \\ \\ d=\sqrt{(2-(-1))^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

The radius is 5.

$(x-2)^2+(y-5)^2=5^2 \\ \boxed{(x-2)^2+(y-5)^2=25}$