Question

Knowing the extreme points of a diameter in circle are `A(-3,1) and D(2,5)`, find the canonical equation of a circle?

Answer 1

`(x+1/2)^2+(y-3)^2=41/4`

Explanation:

The equation of a circle with center `(x_0,y_0)` and radius `r` is

`(x-x_0)^2+(y-y_0)^2=r^2`

To find this equation in the given case, then, we need the radius and the center. As we are given the endpoint of a diameter, and the radius of a circle is half of the length of a diameter, we can find the radius as half of the distance from `A` to `D`.

The distances between two points `(x_1,y_1)` and `(x_2,y_2)` is given by `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`. So we have

`r = 1/2"dist"(A,D) = sqrt((2-(-3))^2+(5-1)^2)/2 = sqrt(41)/2`

Next, we find the center of the circle by using the fact that the midpoint of any diameter of the circle will be its center point. The midpoint of the line segment connecting `(x_1,y_1)` and `(x_2,y_2)` is `((x_1+x_2)/2,(y_1+y_2)/2)`. So we have the center as

`((-3+2)/2,(1+5)/2) = (-1/2,3)`

Now that we have the radius and center, we can plug in our values to find the equation of the circle.

`(x-(-1/2))^2 + (y-3)^2 = (sqrt(41)/2)^2`

or

`(x+1/2)^2+(y-3)^2=41/4`