A circle has a center that falls on the line #y = 1/3x +7 # and passes through # ( 3 ,1 )# and #(7 ,9 )#. What is the equation of the circle?

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Answer 1 · 2017-09-20T20:10:49

The equation of the circle is:

#(x-(-39/5))^2+(y-22/5)^2=(2/5sqrt281)^2#

Use the standard Cartesian form of the equation of a circle

#(x-h)^2+(y-k)^2=r^2#

and the two points to write two equations:

#(3-h)^2+(-6-k)^2=r^2" [1]"#
#(7-h)^2+(2-k)^2=r^2" [2]"#

Expand the squares:

#9-6h+ h^2+36+12k+k^2=r^2" [1.1]"#
#49-14h+h^2+4-4k+k^2=r^2" [2.1]"#

Subtract equation [2.1] from equation [1.1]

#-40+8h+32+16k = 0#

Combine like terms:

#8h+16k = 8#

Divide both sides by 8:

#h+2k = 1" [3]"#

Evaluate the equation #y = 1/3x +7# at the center point #(h,k)#:

#k = 1/3h +7" [4]"#

This allows us to substitute #1/3h + 7# into equation [3]:

#h+2(1/3h + 7) = 1#

#h = -39/5#

Use equation [4] to find the value of k:

#k = 1/3((-39)/5) +7#

#k= 22/5#

Substitute the values of h and k into equation [1]:

#(3-(-39/5))^2+(-6-22/5)^2=r^2#

#r = 2/5sqrt281#

Substitute the values for h, k and r into the standard form:

#(x-(-39/5))^2+(y-22/5)^2=(2/5sqrt281)^2#