A circle has a center that falls on the line #y = 1/8x +4 # and passes through # ( 5 ,8 )# and #(5 ,6 )#. What is the equation of the circle?

Question

1485 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-22T10:21:43

#(x-24)^2+(y-7)^2=362#

Using the two given points #(5, 8)# and #(5, 6)#

Let #(h, k)# be the center of the circle

For the given line #y=1/8x+4#, #(h, k)# is a point on this line.
Therefore, #k=1/8h+4#

#r^2=r^2#

#(5-h)^2+(8-k)^2=(5-h)^2+(6-k)^2#

#64-16k+k^2=36-12k+k^2#

#16k-12k+36-64=0#

#4k=28#

#k=7#

Use the given line #k=1/8h+4#

#7=1/8*h+4#

#h=24#

We now have the center #(h, k)=(7, 24)#

We can now solve for the radius r

#(5-h)^2+(8-k)^2=r^2 #(5-24)^2+(8-7)^2=r^2#

#(-19)^2+1^2=r^2#
#361+1=r^2#

#r^2=362#

Determine now the equation of the circle

#(x-h)^2+(y-k)^2=r^2#
#(x-24)^2+(y-7)^2=362#

The graphs of the circle #(x-24)^2+(y-7)^2=362# and the line #y=1/8x+4#

graph{((x-24)^2+(y-7)^2-362)(y-1/8x-4)=0[-55,55,-28,28]}

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