A circle has a center that falls on the line $y = 6/7x +7$ and passes through $( 7 ,8 )$ and $(3 ,1 )$ . What is the equation of the circle?

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Answer 1 · 2016-04-17T13:06:42

Equation of the circle:

$(x-1/4)^2+(y-101/14)^2=36205/784$

$k=6/7h+7" "$ first equation
$(x-h)^2+(y-k)^2=r^2$

$(7-h)^2+(8-k)^2=r^2" "$ second equation
$(3-h)^2+(1-k)^2=r^2" "$ third equation

We have 3 equations with 3 unknows h, k, r

Using second and third equations, we can eliminate r

$r^2=r^2$
$(7-h)^2+(8-k)^2=(3-h)^2+(1-k)^2$

$49-14h+h^2+64-16k+k^2=9-6h+h^2+1-2k+k^2$
$103-8h-14k=0$

Our fourth equation
$8h+14k=103$

use the first equation now with the fourth equation
$8h+14k=103$
$8h+14(6/7h+7)=103$
$8h+12h+98=103$
$20h=5$
$h=1/4$

solve k:

$k=6/7h+7$
$k=(6/7)(1/4)+7$
$k=3/14+7$
$k=(3+98)/14$
$k=101/14$

solve for r:

$(3-h)^2+(1-k)^2=r^2$
$(3-1/4)^2+(1-101/14)^2=r^2$
$(11/4)^2+(-87/14)^2=r^2$
$121/16+7569/196=r^2$
$r^2=36205/784$

Equation of the circle:

$(x-1/4)^2+(y-101/14)^2=36205/784$

kindly see the graph of line $y=6/7x+7$ and the circle $(x-1/4)^2+(y-101/14)^2=36205/784$

graph{((x-1/4)^2+(y-101/14)^2-36205/784)(y-6/7x-7)=0[-30,30,-15,15]}

God bless....I hope the explanation is useful.

A circle has a center that falls on the line y = 6/7x +7 y = 6/7x +7 and passes through ( 7 ,8 ) ( 7 ,8 ) and (3 ,1 )(3 ,1 ). What is the equation of the circle?