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

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Answer 1 · 2017-06-24T12:15:59

The equation of the circle is $(x-11/18)^2+(y-35/6)^2=2405/162$

Let $C$ be the mid point of $A=(4,4)$ and $B=(1,2)$

$C=((4+1)/2,(4+2)/2)=(5/2,3)$

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

The slope of the line perpendicular to $AB$ is $=-3/2$

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

$y-3=-3/2(x-5/2)$

$y=-3/2x+15/4+3=-3/2x+27/4$

The intersection of this line with the line $y=3x+4$ gives the center of the circle.

$3x+4=-3/2x+27/4$

$3x+3/2x=27/4-4$

$9/2x=11/4$

$x=11/18$

$y=3*(11/18)+4=35/6$

The center of the circle is $(11/18,35/6)$

The radius of the circle is

$r^2=(1-11/18)^2+(2-35/6)^2$

$=(7/18)^2+(-23/6)^2$

$=4810/324=2405/162$

The equation of the circle is

$(x-11/18)^2+(y-35/6)^2=2405/162$
graph{((x-11/18)^2+(y-35/6)^2-2405/162)(y-3x-4)(y+3/2x-27/4)=0 [-9.25, 10.75, -2.56, 7.44]}

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