A circle has a center that falls on the line $y = 4/7x +6$ and passes through $( 2 ,1 )$ and $(3 ,5 )$ . What is the equation of the circle?

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Answer 1 · 2017-06-30T11:49:33

The equation of the circle is $(x+70/23)^2+(y-98/23)^2=19081/529$

Let $C$ be the mid point of $A=(2,1)$ and $B=(3,5)$

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

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

The slope of the line perpendicular to $AB$ is $=-1/4$

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

$y-3=-1/4(x-2)$

$y=-1/4x+1/2+3=-1/4x+7/2$

The intersection of this line with the line $y=4/7x+6$ gives the center of the circle.

$-1/4x+7/2=4/7x+6$

$1/4x+4/7x=7/2-6$

$23/28x=-5/2$

$x=-28/23*5/2=-70/23$

$y=4/7*(-70/23)+6=98/23$

The center of the circle is $(-70/23,98/23)$

The radius of the circle is

$r^2=(2+70/23)^2+(1-98/23)^2$

$=(116/23)^2+(75/23)^2$

$=19081/529$

The equation of the circle is

$(x+70/23)^2+(y-98/23)^2=19081/529$

graph{ ((x+70/23)^2+(y-98/23)^2-19081/529)(y-4/7x-6)(y+1/4x-7/2)=0 [-36.54, 36.55, -18.26, 18.28]}

A circle has a center that falls on the line y = 4/7x +6 y = 4/7x +6 and passes through ( 2 ,1 ) ( 2 ,1 ) and (3 ,5 )(3 ,5 ). What is the equation of the circle?