A circle has a center that falls on the line #y = 3/7x +1 # and passes through # ( 2 ,8 )# and #(3 ,5 )#. What is the equation of the circle?

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Answer 1 · 2016-10-03T23:15:36

The equation is:

#2405 = (x - 49)² + (x - 22)²#

#r² = (x - h)² + (y - k)²#

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

Using the two given points, we write two equations:

#r² = (2 - h)² + (8 - k)²#
#r² = (3 - h)² + (5 - k)²#

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

#(2 - h)² + (8 - k)²= (3 - h)² + (5 - k)²#

Using the pattern, #(a - b)² = a² - 2ab + b²# we expand the squares:

#4 - 4h + h² + 64 - 16k + k²= 9 - 6h + h² + 25 - 10k + k²#

#4 - 4h + 64 - 16k= 9 - 6h + 25 - 10k#

#4 + 64 -6k= 9 - 2h + 25#

#-6k = -2h - 34#

#k = 1/3h + 17/3#

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

#k = 3/7h + 1#

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

#3/7h + 1 = 1/3h + 17/3#

#9h + 21 = 7h + 119#

#2h = 98#

#h = 49#

#k = 3/7(49) + 1#

#k = 22#

Substitute the center into the first equation and solve for r²:

#r² = (2 - 49)² + (8 - 22)²#

#r² = (-47)² + (-14)²#

#r² = 2405#

The equation is:

#2405 = (x - 49)² + (x - 22)²#