A circle has a center that falls on the line #y = 1/7x +4 # and passes through # ( 7 ,8 )# and #(3 ,6 )#. What is the equation of the circle?

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Answer 1 · 2018-04-23T15:36:58

#color(blue)((x-91/15)^2+(y-73/15)^2=481/45)#

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

Where #(h,k)# are the #x and y# coordinates of the centre respectively and #r# is the radius.

The centre lie on the line #y=1/7x+4#

#:.#

#k=1/7h+4 \ \ \ \ \ \ \[1]#

#(7,8) and (3,6)# lie on the circle:

#(7-h)^2+(8-k)^2=r^2 \ \ \ \ \ \ \[2]#

#(3-h)^2+(6-k)^2=r^2 \ \ \ \ \ \ \[3]#

Subtract #[3]# from #[2]#

#68-8h-4k=0#

Substituting #k=1/7h+4#

#68-8h-4(1/7h+4)=0#

#52-60/7h=0=>h=91/15#

Substituting in #[1]#:

#k=1/7(91/15)+4#

#k=73/15#

We now need the value of #r^2#

Using coordinate #(3,6)#

#(3-91/15)^2+(6-73/15)^2=r^2#

#r^2=481/45#

So the equation of the circle is:

#color(blue)((x-91/15)^2+(y-73/15)^2=481/45)#