A circle has a center that falls on the line $y = 2/7x +3$ and passes through $( 2 ,4 )$ and $(6 ,5 )$ . What is the equation of the circle?

Question

1447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-19T12:26:59

The equation of the circle is $(x-49/12)^2+(y-25/6)^2=629/144$

Let $C$ be the mid point of $A=(2,4)$ and $B=(6,5)$

$C=((2+6)/2,(4+5)/2)=(4,9/2)$

The slope of $AB$ is $=(5-4)/(6-2)=1/4$

The slope of the line perpendicular to $AB$ is $=-4$

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

$y-9/2=-4(x-4)$

$y=-4x+16+9/2=-4x+41/2$

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

$2/7x+3=-4x+41/2$

$2/7x+4x=41/2-3$

$30/7x=35/2$

$x=35/2*7/30=245/60=49/12$

$y=2/7*245/60+3=25/6$

The center of the circle is $(49/12,25/6)$

The radius of the circle is

$r^2=(49/12-2)^2+(25/6-4)^2$

$=(25/12)^2+(1/6)^2$

$=629/144$

The equation of the circle is

$(x-49/12)^2+(y-25/6)^2=629/144$
graph{((x-49/12)^2+(y-25/6)^2-629/144)(y-2/7x-3)=0 [-3.146, 9.34, 0.674, 6.914]}

A circle has a center that falls on the line y = 2/7x +3 y = 2/7x +3 and passes through ( 2 ,4 ) ( 2 ,4 ) and (6 ,5 )(6 ,5 ). What is the equation of the circle?