A circle has a center that falls on the line $y = 3/8x +3$ and passes through $( 1 ,4 )$ and $(2 ,9 )$ . What is the equation of the circle?

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Answer 1 · 2017-06-22T10:12:01

The equation of the circle is $(x-152/23)^2+(y-126/23)^2=17797/529$

Let $C$ be the mid point of $A=(1,4)$ and $B=(2,9)$

$C=((1+2)/2,(9+4)/2)=(3/2,13/2)$

The slope of $AB$ is $=(9-4)/(2-1)=(5)/(1)=5$

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

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

$y-13/2=-1/5(x-3/2)$

$y=-1/5x+3/10+13/2=-1/5x+34/5$

The intersection of this line with the line $y=3/8x+3$ gives the center of the circle.

$3/8x+3=-1/5x+34/5$

$3/8x+1/5x=34/5-3$

$23/40x=19/5$

$x=19/5*40/23=152/23$

$y=3/8*152/23+3=126/23$

The center of the circle is $(152/23,126/23)$

The radius of the circle is

$r^2=(1-152/23)^2+(4-126/23)^2$

$=(129/23)^2+(-34/23)^2$

$=17797/529$

The equation of the circle is

$(x-152/23)^2+(y-126/23)^2=17797/529$
graph{((x-152/23)^2+(y-126/23)^2-17797/529)(y-3/8x-3)=0 [-8.25, 20.23, -2.34, 11.9]}

A circle has a center that falls on the line y = 3/8x +3 y = 3/8x +3 and passes through ( 1 ,4 ) ( 1 ,4 ) and (2 ,9 )(2 ,9 ). What is the equation of the circle?