A circle has a center that falls on the line #y = 2/9x +8 # and passes through # ( 2 ,5 )# and #(1 ,7 )#. What is the equation of the circle?

1 Answer
Apr 15, 2016

Explanation:

Let the coordinate of the centre of the circle be (a,b) and radius is r.
Then its equation will be
#(x-a)^2+(y-b)^2=r^2#

The circle passes through (2,5) and (1,7) show these points will satisfy the equation of the circle, hence

#(2-a)^2+(5-b)^2=r^2.......(1)#
#(1-a)^2+(7-b)^2=r^2.......(2) #
Subtracting (2) from (1) we have

#(2-a)^2-(1-a)^2+(5-b)^2-(7-b)^2=0#

#=>(2-a-1+a)(3-2a)+(5-b-7+b)(12-2b)=0#
#=>3-2a-24+4b=0#
#=>2a-4b=-21......(3)#

Again The center (a,b) lies on the given line #y=2/9x+8# So we can write #b=2/9a+8......(4)#

From (3) and (4)
we have
#=>2a-4(2/9a+8)=-21#

#=>2a-8/9a-32=-21#
#=>2a-8/9a=11#
#=>10a=99#
#:,a=9.9#
From (4)
#b=2/9a+8#
#=>b=2/9xx9.9+8=10.2#
From eq (1)
#(2-9.9)^2+(5-10.2)^2=r^2.......(1)#
#=>r^2=62.41+27.04=89.45#

Equation of the Circle
#(x-9.9)^2+(y-10.2)^2=89.45^2#