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?

Question

1324 views around the world

You can reuse this answer
Creative Commons License

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.