A circle has a center that falls on the line $y = 1/7x +4$ and passes through $( 7 ,8 )$ and $(3 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2016-12-27T22:29:19

The equation of a circle with center $(x_o,y_o)$ and radius $R$ is

$(x-x_o)^2+(y-y_o)^2=R^2$

Hence the center belongs to the line we have that

$y_o=1/7*x_o +4$

Because the circle passes from point $(7,8)$ and $(3,1)$ we have also

$(7-x_o)^2+(8-y_o)^2=R^2$

and

$(3-x_o)^2+(1-y_o)^2=R^2$

Hence the RHS of both equations is the same ( $=R^2$ )
we have

$(7-x_o)^2+(8-y_o)^2=(3-x_o)^2+(1-y_o)^2$

$7^2-14*x_o + (x_o)^2+8^2-16*y_o + (y_o)^2= 3^2-6*x_o + (x_o)^2+1-2*y_o + (y_o)^2$

$49+64-14*x_o-16*y_o-9+6*x_o-1+2*y_o=0$

$-8*x_o-14*y_o + 103=0$

$8*x_o + 14*y_o - 103=0$

Because $y_o=1/7*x_o +4$ and $8*x_o + 14*y_o - 103=0$

solving the system of these two equation yields the center of the

circle

$(x_o,y_o)=(47/10,327/70)$

and the radius is

$(3-47/10)^2+(1-327/70)^2=R^2$

$R=sqrt(8021/490)$

Finally the equation of the circle is

$(x-47/10)^2+(y-327/70)^2=8021/490$

A circle has a center that falls on the line y = 1/7x +4 y = 1/7x +4 and passes through ( 7 ,8 ) ( 7 ,8 ) and (3 ,1 )(3 ,1 ). What is the equation of the circle?