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# Circle

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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
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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

$$d=\sqrt{(-1-7)^2+(2-8)^2}$$

$$d=\sqrt{(-8)^2+(-6)^2}$$

$$d=\sqrt{64+36}$$

$$d=\sqrt{100}$$

$$d=10$$

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
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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   Sep 16, 2019