Find an equation of the circle that satisfies the given conditions.

Endpoints of a diameter are P(−1, 2) and Q(7, 8)

mharrigan920 Sep 16, 2019

#1**+1 **

The form of the equation of a circle is r^{2}=(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}

power27 Sep 16, 2019

#2**+3 **

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

CPhill Sep 16, 2019