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

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

 Sep 16, 2019
 #1
avatar+196 
+1

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

 Sep 16, 2019
 #2
avatar+104876 
+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

 

 

cool cool cool

 Sep 16, 2019

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