Find an equation of the circle that satisfies the given conditions.
Endpoints of a diameter are P(−1, 2) and Q(7, 8)
The form of the equation of a circle is r2=(x-a)2+(y-b)2, where r is the radius and (a,b) is the center.
The length of the line PQ is the diamiter of the circle, so plug (-1,2) and (7,8) into the distance formula
\(d=\sqrt{(-1-7)^2+(2-8)^2}\)
\(d=\sqrt{(-8)^2+(-6)^2}\)
\(d=\sqrt{64+36}\)
\(d=\sqrt{100}\)
\(d=10\)
so the radius is 5
The midpoint of the line PQ is the center of the circle, so plug P and Q into the midpoint formula
\(m=(\frac{-1+7}{2},\frac{8+2}{2})\)
\(m=(\frac{6}{2},\frac{10}{2})\)
m=(3,5)
Plugging the midpoint and the radius into the equation for the circle gets:
25=(x-3)2+(y-5)2
P(−1, 2) and Q(7, 8)
The midpoint of PQ will be the center of the ciricle
So
( [ -1 + 7 ] /2 , [ 8 + 2 ] /2 ) = [ 6/2 , 10/2) = (3 , 5) = (h, k)
And the distance from this point to either P or Q is the radius, r
And we can find r^2 as
(7 -3)^2 + ( 8 - 5)^2 = r^2
4^2 + 3^2 = r^2
16 + 9 = r^2
25 = r^2
So....the equation of the circle is
(x - h)^2 + (y - k)^2 = r^2
( x - 3)^2 + (y - 5)^2 = 25