A circle has a center that falls on the line #y = 2/7x +3 # and passes through # ( 1 ,4 )# and #(6 ,1 )#. What is the equation of the circle?
1 Answer
#(x-133/29)^2+(y-125/29)^2 = 10897/29^2#
Or if you prefer:
#(29x-133)^2+(29y-125)^2 = 10897#
Explanation:
Since the circle passes through
Let
The midpoint of
The slope of
So the slope of any line perpendicular to
So the equation of the line through the midpoint of
#(y - 5/2) = 5/3(x-7/2)#
Add
#y = 5/3x - 10/3#
Find the intercept of this line and the given line by equating the values of
#5/3x - 10/3 = y = 2/7x + 3#
Multiply both ends by
#35x - 70 = 6x + 63#
Add
#35x = 6x + 133#
Subtract
#29x = 133#
Hence
So:
#y = 5/3x - 10/3 = 5/3*133/29 - 10/3 = (665-290)/57 = 375/57 = 125/29#
So the equation of our circle can be written in the form:
#(x-133/29)^2+(y-125/29)^2 = r^2#
Since the circle passes through
#r^2 = (1-133/29)^2+(4-125/29)^2#
#=(-104/29)^2+(-9/29)^2#
#=(104^2+9^2)/29^2#
#=(10816+81)/29^2#
#=10897/29^2#
So the equation of the circle may be written:
#(x-133/29)^2+(y-125/29)^2 = 10897/29^2#
graph{((x-133/29)^2+(y-125/29)^2-10897/29^2)((x-1)^2+(y-4)^2-0.02)((x-6)^2+(y-1)^2-0.02)(y - 2/7x-3)((x-133/29)^2+(y-125/29)^2-0.04)=0 [-6.58, 13.42, -1, 9]}