A circle has a center that falls on the line #y = 1/3x +7 # and passes through # ( 3 ,1 )# and #(6 ,9 )#. What is the equation of the circle?

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Answer 1 · 2016-06-23T03:11:30

#color(red)((x--15/34)^2+(y-233/34)^2=26645/578)#

For the first point #(3, 1)# we can set up one equation

#(x-h)^2+(y-k)^2=r^2#

#(3-h)^2+(1-k)^2=r^2#

For the second point #(6, 9)# we can set up another equation

#(x-h)^2+(y-k)^2=r^2#

#(6-h)^2+(9-k)^2=r^2#
~~~~~~~~~~~~~~~~~~~~~~~~~~
#r^2=r^2#
#(3-h)^2+(1-k)^2=(6-h)^2+(9-k)^2#

After simplification, we now have an equation

#6h+16k=107#

Also from the line #y=x/3+7# which contains the point #(h, k)#, we have

#k=h/3+7# or #h-3k=-21#

Simultaneous solution of the system
#h-3k=-21#
#6h+16k=107#

results to a center #(h, k)=(-15/34, 233/34)#

the radius can be computed now using center #(h, k)=(-15/34, 233/34)# and #(6-h)^2+(9-k)^2=r^2#

#(6-h)^2+(9-k)^2=r^2#
#r^2=(6--15/34)^2+(9-233/34)^2#
#r^2=26645/578#

We can now form the equation of the circle

#(x-h)^2+(y-k)^2=r^2#

#color(red)((x--15/34)^2+(y-233/34)^2=26645/578)#

Kindly see the graph for visual inspection
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God bless....I hope the explanation is useful.