A circle has a center that falls on the line $y = 2/7x +3$ and passes through $( 2 ,8 )$ and $(3 ,5 )$ . What is the equation of the circle?

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Answer 1 · 2017-06-30T06:40:46

The equation of the circle is $(x+56)^2+(y+13)^2=3805$

Let $C$ be the mid point of $A=(2,8)$ and $B=(3,5)$

$C=((2+3)/2,(8+5)/2)=(5/2,13/2)$

The slope of $AB$ is $=(5-8)/(3-2)=(-3)/(1)=-3$

The slope of the line perpendicular to $AB$ is $=1/3$

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

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

$y=1/3x-5/6+13/2=1/3x+17/3$

The intersection of this line with the line $y=2/7x+3$ gives the center of the circle.

$1/3x+17/3=2/7x+3$

$1/3x-2/7x=3-17/3$

$1/21x=-8/3$

$x=-21*8/3=-56$

$y=2/7*(-56)+3=-13$

The center of the circle is $(-56,-13)$

The radius of the circle is

$r^2=(2+56)^2+(8+13)^2$

$=(58)^2+(21)^2$

$=3805$

The equation of the circle is

$(x+56)^2+(y+13)^2=3805$

graph{((x+56)^2+(y+13)^2-3805)(y-2/7x-3)(y-1/3x-17/3)=0 [-165.8, 101, -73.3, 60.2]}

A circle has a center that falls on the line y = 2/7x +3 y = 2/7x +3 and passes through ( 2 ,8 ) ( 2 ,8 ) and (3 ,5 )(3 ,5 ). What is the equation of the circle?