If $A = (3,4)$ and $C = (-7,2)$ are opposite vertices of a rectangle $ABCD,$ then vertices $B$ and $D$ must lie on the circle
\[x^2 + y^2 - px - qy + s = 0.\]
Compute the ordered triple of real numbers $(p,q,s).$
+129771 B= (-7,4) C = (3,2)
The midoint of BD = M = [ ( -7 + 3) / 2 , (2 + 4) /2] = (-2, 3) = the center of the circle
The distance from B to M = sqrt [ (-7 - -2)^2 + (4 -3)^2 ] = sqrt [ 25 + 1] = sqrt (26) = radius of the circle
So we have
(x - -2)^2 + ( y - 3)^2 = 26
( x + 2)^2 + ( y -3)^2 = 26
x^2 + 4x + 4 + y^2 - 6y + 9 = 26
x^2 + y^2 + 4x - 6y - 13 = 0
x^2 + y^2 - (-4)x - 6y + (-13) = 0
(p , q, s) = (-4, 6, -13)
