A circle has a center that falls on the line $y = 1/7x +7$ and passes through $( 7 ,8 )$ and $(3 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2017-06-21T07:49:30

The equation of the circle is
$(x-1/2)^2+(y-99/14)^2=4225/98$

Let $C$ be the mid point of $A=(7,8)$ and $B=(3,1)$

$C=((7+3)/2,(8+1)/2)=(5,9/2)$

The slope of $AB$ is $=(1-8)/(3-7)=(-7)/(-4)=7/4$

The slope of the line perpendicular to $AB$ is $=-4/7$

The equation of the line passing trrough $C$ and perpendicular to $AB$ is

$y-9/2=-4/7(x-5)$

$y=-4/7x+20/7+9/2=-4/7x+103/14$

The intersection of this line with the line $y=1/7x+7$ gives the center of the circle.

$1/7x+7=-4/7x+103/14$

$1/7x+4/7x=103/14-7$

$5/7x=5/14$

$x=5/14*7/5=1/2$

$y=1/7*1/2+7=99/14$

The center of the circle is $(1/2,99/14)$

The radius of the circle is

$r^2=(3-1/2)^2+(1-99/14)^2$

$=(5/2)^2+(-85/14)^2$

$=8450/196=4225/98$

The equation of the circle is

$(x-1/2)^2+(y-99/14)^2=4225/98$
graph{ ((x-1/2)^2+(y-99/14)^2-4225/98)(y-1/7x-7)=0 [-20.02, 20.52, -2.84, 17.44]}

A circle has a center that falls on the line y = 1/7x +7 y = 1/7x +7 and passes through ( 7 ,8 ) ( 7 ,8 ) and (3 ,1 )(3 ,1 ). What is the equation of the circle?