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?

Question

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?

Ella Armstrong · Accepted Answer

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#

Allison Allen · Answer

undefined The equation of the circle is $ (x - 5)^2 + (y - 5)^2 = 10 $.