A circle has a center that falls on the line #y = 3x +4 # and passes through #(4 ,4 )# and #(9 ,2 )#. What is the equation of the circle?

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Answer 1 · 2017-05-10T06:48:31

#x^2+y^2+69x+199y=1104#

The midpoint of segment joining #(4,4)# and #(9,2)# is #((4+9)/2,(4+2)/2)# i.e. #(13/2,3)#

Further slope of this segment is #(2-4)/(9-4)=(-2)/5# and slope of line perpendicular to it would be #(-1)/((-2)/5)=5/2#.

Hence, equation of the perpendicular bisector (on which we have the centre as it is equidistant from the two given points) is

#(y-3)=5/2(x-13/2)# or #4y-12=10x-65# or #10x-4y=53#

Solving #10x-4y=53# and #y=3x+4# gives us centre of the circle

#10x-4(3x+4)=53# or #-2x=69# or #x=-69/2#

and #y=3xx(-69/2)+4=-199/2# and hence centre is #(-69/2,-199/2)#

Its distance from #(4,4)# is radius and hence equation of circle is

#(x+69/2)^2+(y+199/2)^2=(4+69/2)^2+(4+199/2)^2#

or #x^2+69x+y^2+199y=4^2+276+4^2+796#

or #x^2+y^2+69x+199y=1104#