Question

How do you write the equation of the circle with a diameter that has endpoints at (8, 7) and (–4, –3)?

Answer 1

Find the center and radius of the circle, and then write it in standard form as

`(x-2)^2+(y-2)^2=61`

Explanation:

The standard equation of a circle centered at `(x_0,y_0)` with radius `r` is `(x-x_0)^2+(y-y_0)^2=r^2`. Thus, if we can find the center and the radius, we can write the equation.

As the midpoint of any diameter of a circle is the center of the circle, we can find the center by locating the midpoint of the line segment with endpoints `(8,7)` and `(-4,-3)`. The midpoint of between any two points `(a_1,b_1)` and `(a_2,b_2)` is `((a_1+a_2)/2,(b_1+b_2)/2)`. Then, substituting our values, we have the midpoint, and thus the circle's center, at

`((8-4)/2,(7-3)/2)=(2,2)`

Next, to find the radius of the circle, we can just calculate the distance from the center to either of the given points on the circle. The distance between two points `(a_1,b_1),(a_2,b_2)` is `sqrt((a_2-a_1)^2+(b_2-b_1)^2)`. Then, using the center at the point `(8,7)`, we have the radius as

`sqrt((8-2)^2+(7-2)^2)=sqrt(36+25)=sqrt(61)`

With both the center at the radius, we can now write the circle's equation:

`(x-2)^2+(y-2)^2=(sqrt(61))^2`

`:. (x-2)^2+(y-2)^2=61`