A circle has a center that falls on the line $y = 1/7x +4$ and passes through $( 5 ,8 )$ and $(5 ,6 )$ . What is the equation of the circle?

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Answer 1 · 2017-02-18T11:31:37

The equation of the circle is $(x-21)^2+(y-7)^2=257$

A line passing through the mid point of $(5,8)$ and $(5,6)$ and

parallel to the x-axis will cut the line $y=x/7+4$ at the center of

the circle.

Let $(a,b)$ be the center of the circle

$7y=7=a/7+4$

$a/7=7-4=3$

$a=21$ and $b=7$

The center is $(21,7)$

The radius of the circle is

$r=sqrt((21-5)^2+(7-8)^2)$

$=sqrt(16^2+1)$

$=sqrt257$

The equation of the circle is

$(x-21)^2+(y-7)^2=257$

graph{((x-21)^2+(y-7)^2-257)(y-x/7-4)(y-7)=0 [-26.26, 46.76, -9.13, 27.44]}

A circle has a center that falls on the line y = 1/7x +4 y = 1/7x +4 and passes through ( 5 ,8 ) ( 5 ,8 ) and (5 ,6 )(5 ,6 ). What is the equation of the circle?