A circle has a center that falls on the line #y = 7/9x +7 # and passes through # ( 2 ,5 )# and #(5 ,1 )#. What is the equation of the circle?

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Answer 1 · 2016-10-02T18:29:52

The equation of the circle is:

#91512.5 = (x - -477/2)² + (y - -357/2)²#

#r² = (x - h)² + (y - k)²#

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

We can use the above to write two equations, using the given points:

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

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

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

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

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

Combine like terms:

#3 + 6h = 8k#

Flip and divide both sides by 8:

#k = 3/4h + 3/8#

Substitute (h, k) into the given equation of the line on which the center lies:

#k = 7/9h + 7#

Because #k = k#, we can use the right sides of both linear equations for the center to solve for h:

#7/9h + 7 = 3/4h + 3/8#

Multiply both sides of the equation by 72:

#56h + 504 = 54h + 27#

#2h = -477#

#h = -477/2#

#k = -357/2#

Using the point (2, 5) and the center #(-477/2, -357/2)# we compute the square of the radius:

#r² = (2 - -477/2)² + (5 - -357/2)²#

#r² = 91512.5#

The equation of the circle is:

#91512.5 = (x - -477/2)² + (y - -357/2)²#