A circle has a center that falls on the line $y = 5/8x +1$ and passes through $( 5 ,2 )$ and $(3 ,4 )$ . What is the equation of the circle?

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Answer 1 · 2017-06-10T11:59:42

The equation of the circle is $(x-16/3)^2+(y-13/3)^2=50/9$

Let $C$ be the mid point of $A=(5,2)$ and $B=(3,4)$

$C=((5+3)/2,(2+4)/2)=(4,3)$

The slope of $AB$ is $=(4-2)/(3-5)=-2/2=-1$

The slope of the line perpendicular to $AB$ is $=1$

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

$y-3=1(x-4)$

$y=x-1$

The intersection of this line with the line $y=5/8x+1$ gives the center of the circle.

$5/8x+1=x-1$

$3/8x=2$

$x=16/3$

$y=16/3-1=13/3$

The center of the circle is $(16/3,13/3)$

The radius of the circle is

$r^2=(16/3-5)^2+(13/3-2)^2$

$=(1/3)^2+(7/3)^2$

$=50/9$

The equation of the circle is

$(x-16/3)^2+(y-13/3)^2=50/9$

graph{((x-16/3)^2+(y-13/3)^2-50/9)(y-5/8x-1)(y-x+1)=0 [0.604, 9.372, 1.62, 6.002]}

A circle has a center that falls on the line y = 5/8x +1 y = 5/8x +1 and passes through ( 5 ,2 ) ( 5 ,2 ) and (3 ,4 )(3 ,4 ). What is the equation of the circle?