A circle has a center that falls on the line #y = 5/8x +6 # and passes through # ( 1 ,5 )# and #(2 ,4 )#. What is the equation of the circle?

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Answer 1 · 2016-10-03T16:19:58

#2384/361 = (x - 24/19)² + (y - 148/19)²#

#r² = (x - h)² + (y - k)²#

where r is the radius and (h, k) is the center.

Using the given points we can use the above to create two equation:

#r² = (1 - h)² + (5 - k)²#
#r² = (2 - h)² + (4 - k)²#

Because #r² = r²# we can set the right sides equal:

#(1 - h)² + (5 - k)² = (2 - h)² + (4 - k)²#

Use the pattern #(a - b)² = a² - 2ab + b²# to expand the squares:

#1 - 2h + h² + 25 - 10k + k² = 4 - 8h + h² + 16 - 8k + k²#

Combine like terms:

#6h + 6 = 2k#

#k = 3h + 3#

Substitute the point (h, k) into the given linear equation:

#k = 5/8h + 6#

Because k = k we can set the right sides equal:

#3h + 3 = 5/8h + 6#

#24h + 24 = 5h + 48#

#h = 24/19#

#k = 5/8(24/19) + 6 #

#k = 120/152 + 1064/152 #

#k = 1184/152#

#k = 148/19#

#r² = (1 - 24/19)² + (5 - 148/19)²#

#r² = (-5/19)² + (-53/19)²#

#r² = 2384/361#