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

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Answer 1 · 2017-06-24T11:51:27

The equation of the circle is $(x+70/27)^2+(y-122/27)^2=24401/729$

Let $C$ be the mid point of $A=(2,1)$ and $B=(3,6)$

$C=((2+3)/2,(6+1)/2)=(5/2,7/2)$

The slope of $AB$ is $=(6-1)/(3-2)=(5)/(1)=5$

The slope of the line perpendicular to $AB$ is $=-1/5$

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

$y-7/2=-1/5(x-5/2)$

$y=-1/5x+1/2+7/2=-1/5x+4$

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

$4/7x+6=-1/5x+4$

$4/7x+1/5x=4-6$

$27/35x=-2$

$x=-70/27$

$y=4/7*(-70/27)+6=122/27$

The center of the circle is $(-70/27,122/27)$

The radius of the circle is

$r^2=(2+70/27)^2+(1-122/27)^2$

$=(124/27)^2+(-95/27)^2$

$=24401/729$

The equation of the circle is

$(x+70/27)^2+(y-122/27)^2=24401/729$
graph{((x+70/27)^2+(y-122/27)^2-24401/729)(y-4/7x-6)(y+1/5x-4)=0 [-15.77, 12.7, -2.41, 11.83]}

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 ,6 )(3 ,6 ). What is the equation of the circle?