A circle has a center that falls on the line #y = 3x +4 # and passes through #(4 ,4 )# and #(9 ,4 )#. What is the equation of the circle?

1 Answer

Equation of the circle
#(x-13/2)^2+(y-47/2)^2=1546/4#

Explanation:

Let the center #(h,k)#

From the given line #y=3x+4#
we can write #k=3h+4# because #(h, k)# is on the line

Using the given points (4, 4) and (9, 4)

radius=radius

#sqrt((4-h)^2+(4-k)^2)=sqrt((9-h)^2+(4-k)^2)#

also

#(4-h)^2+(4-k)^2=(9-h)^2+(4-k)^2#

#16-8h+h^2=81-18h+h^2#

#18h-8h=81-16#
#10h=65#
#h=13/2#

Using now the equation #k=3h+4#

#k=3(13/2)+4#

#k=39/2+4#

#k=(39+8)/2#

#k=47/2#

the center of the circle
#(h, k)=(13/2, 47/2)#

compute the radius #r# now

#r^2=(4-h)^2+(4-k)^2#

#r^2=(4-13/2)^2+(4-47/2)^2#

#r^2=((8-13)/2)^2+((8-47)/2)^2#

#r^2=(-5/2)^2+(-39/2)^2#

#r^2=1546/4#

We can now obtain the equation of the circle

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

#(x-13/2)^2+(y-47/2)^2=1546/4#

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