A circle has a center that falls on the line #y = 5/4x +8 # and passes through # ( 4 ,7 )# and #(2 ,5 )#. What is the equation of the circle?

1 Answer
Shiva Prakash M V
Mar 7, 2018

center is #-=(4/9,8 5/9)#
#(x-4/9)^2+y-8 5/9)^2=(3 5/9)^2+(1 5/9)^2#
or
#-=(0.444,8.555)#
#(x-0.444)^2+(y-8.555)^2=15.06

Explanation:

Let
#(x,y)#
be the center of the circle
#(4,7)#
and
#(2,5)#
lie on the perimeter.
#(4,7) and (2,5)#
are equidistant from (x,y)
#(x-4)^2+(y-7)^2=(x-2)^2+(y-5)^2#
#x^2-8x+16+y^2-14y+49=x^2-4x+4+y^2-10y+25#
#-8x-14y+16+49=-4x-10y+4+25#
#-8x-14y+65=-4x-10y+29#
#8x-4x+14y-10y+29-65=0#
#4x+4y-36=0#
#x+y=9#
Center lies on the line
#y=5/4x+8#
#x+5/4x+8=9#
#9/4x=9-8#
#9/4x=1#
#x=4/9#
#x+y=9#
#4/9+y=9#
#4/9+y=9#
#y=9-4/9#
#y=8 5/9#
#y=5/4x+8#
#8 5/9=5/4(4/9)+8#
#8 5/9=5/9)+8#
#8 5/9=8 5/9#
lhs=rhs
#(x-4)^2+(y-7)^2=(x-2)^2+(y-5)^2#
#(4/9-4)^2+(8 5/9-7)^2=(4/9-2)^2+(8 5/9-5)^2#
#(3 5/9)^2+(1 5/9)^2=(1 5/9)^2+(3 5/9)^2#
lhs=rhs
and the radius is
#r=sqrt((3 5/9)^2+(1 5/9)^2)#
#r=3.88#
center is #-=(4/9,8 5/9)#

#(x-4/9)^2+y-8 5/9)^2=(3 5/9)^2+(1 5/9)^2#
#-=(0.444,8.555)#
#(x-0.444)^2+(y-8.555)^2=15.06

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