Question

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

Endpoints of a diameter are P(-1, 1) and Q(5,9)

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:

`((x_1+x_2)/(2), (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 \\ \\ (x_1+x_2)/(2)=(-1+5)/(2)=(4)/(2)=2 \\ (y_1+y_2)/(2)=(1+9)/(2)=(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=√((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=√((2-(-1))^2+(5-1)^2)=√(3^2+4^2)=√(9+16)=√(25)=5`

The radius is 5.


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