A circle has a center that falls on the line y = 5/3x +1 and passes through (8 ,2 ) and (3 ,1 ). What is the equation of the circle?

1 Answer
Narad T.
May 27, 2017

The equation of the circle is (x-21/5)^2+(y-8)^2=1261/25

Explanation:

Let the points A=(8,2) and B=(3,1)

The midpoint of AB is C=((8+3)/2, (1+2)/2)=(11/2,3/2)

The slope of AB is m=(1-2)/(3-8)=-1/-5=1/5

The slope of the line perpendicular to AB is m'=-5, as

mm'=-1

The equation of the line, passing through C and perpendicular to AB is L'

y-3/2=-5(x-11/2)

y-3/2=-5x+55/2

y=-5x+55/2+3/2=-5x+29

Let the line y=5/3x+1 be L

The center of the circle O lies at the intersection of L and L'

We have 2 equations

y=5/3x+1 and y=-5x+29

So,

5/3x+1=-5x+29

5/3x+5x=29-1=28

20/3x=28

x=28/20*3=21/5

y=5/3*21/5+1=8

The center of the circle is O=(21/5,8)

The radius of the circle is

r^2=(21/5-3)^2+(8-1)^2=36/25+49=1261/25

The equation of the circle is

(x-21/5)^2+(y-8)^2=1261/25 graph{(y-5/3x-1)(y+5x-29)((x-21/5)^2+(y-8)^2-1261/25)=0 [-18.56, 21.97, -2.56, 17.72]}

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