Find an equation of the circle whose diameter has endpoints ( −1,−4) and (3,2) .

The equation of a circle is:

(x-h)^2+(y-k)^2=r^2, where (h,k) is the center and r is the radius

So the first thing we can do is to find the center of the circle, which will be at the midpoint of the two endpoints of the diameter.  The midpoint is simply the average of the coordinates of the endpoints.

mp=((-1+3)/2, (-4+2)/2)

mp=(1, -1), and this midpoint is the center of the circle our (h,k) for the circle equation.

Now we must find the length of the diameter so that we can know what our radius is.  The distance between any two points is:

d^2=(x2-x1)^2+(y2-y1)^2 so

d^2=(-1-3)^2+(-4-2)^2

d^2=4^2+6^2

d^2=16+36

d^2=52

Note that this distance squared is our diameter squared.

Since d=2r

d^2=4r^2

r^2=d^2/4, we showed that d^2=52 so

r^2=52/4 

r^2=13  (r=√13, but we need r^2 for our equation anyway)

So now we can fill out our circle equation because we know the center is at (1,-1) are r^2=13

(x-1)^2+(y+1)^2=13

Multiply 9 and 67. then multiply 3 with 67. lastly the answer that you got for 9x67 is to be subtracted by the answer you got for 3x67