How do you write the equation of the circle with diameter that has endpoints at (–4, –1) and (–8, –9)?

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Answer 1 · 2018-06-23T09:42:57

$(x+6)^2+(y+5)^2=20$

To write the equation of a circle we need to know the center and the radius. Since we know two endpoints, we know the extrema of a diameter.

The idea is the following: we will find the center as the midpoint of the diameter, and the radius as half the length of the diameter.

To find the midpoint of a segmente with extrema $A=(x_A,y_A)$ and $B=(x_B,y_B)$ , we have the following formula:

$M = (\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2})$

In your case, this becomes

$C = (\frac{-4-8}{2}, \frac{-1-9}{2})=(-6,-5)$

As for the length, we have the following formula:

$d = sqrt((x_A-x_B)^2+(y_A-y_B)^2)$

Which in your case becomes

$d = sqrt((-4+8)^2+(-1+9)^2)=sqrt(16+64)=sqrt(80) = 4sqrt(5)$

So, the diameter is $4sqrt(5)$ units long, which means that the radius is $2sqrt(5)$ units long.

Now we can write the equation:

in general, given the center $(x_0,y_0)$ and the radius $r$ , we have

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

Which in this case becomes

$(x+6)^2+(y+5)^2=20$