A circle has a center that falls on the line #y = 1/4x +2 # and passes through # ( 3 ,7 )# and #(7 ,1 )#. What is the equation of the circle?

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Answer 1 · 2017-11-23T04:44:29

#(x - 16/5)^2+(y-14/5)^2 = (sqrt442/5)^2#

The Cartesian form for the equation of a circle is:

#(x - h)^2+(y-k)^2 = r^2" [1]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center, and #r# is the radius.

We can use the two points, #(3,7)# and #(7,1)# and equation [1] to write two equations:

#(3 - h)^2+(7-k)^2 = r^2" [2]"#
#(7 - h)^2+(1-k)^2 = r^2" [3]"#

Expand the squares:

#9 - 6h+h^2+49-14k+k^2 = r^2" [2.1]"#
#49 - 14h+h^2+1-2k+k^2 = r^2" [3.1]"#

Subtract equation [3.1] from equation [2.1]:

#8h-12k+8=0" [4]"#

Evaluate the given line at the point #(h,k)#:

#k = 1/4h+2" [5]"#

Substitute equation [5] into equation [4]:

#8h-12(1/4h+2)+8=0#

#8h-3h-24+8=0#

#5h = 16#

#h = 16/5#

Use equation [5] to solve for k:

#k = 1/4(16/5)+2#

#k = 14/5#

Use equation [2] to find the value of r:

#(3 - 16/5)^2+(7-14/5)^2 = r^2#

#(-1/5)^2+(21/5)^2 = r^2#

#1/25+441/25 = r^2#

#r^2= 442/25#

#r = sqrt442/5#

Substitute the know values into equation [1]:

#(x - 16/5)^2+(y-14/5)^2 = (sqrt442/5)^2#