A circle has a center that falls on the line #y = 7/9x +5 # and passes through # ( 7 ,3 )# and #(5 ,1 )#. What is the equation of the circle?

Answer 1

The equation of the circle is #(x-27/16)^2+(y-101/16)^2=39.2#

Let the center of the circle be #(a,b)# Then the equation of the circle is #(x-a)^2+(y-b)^2=r^2# As the circles passes through #(7,3)# and #(5,1)# Then substituting those points in the equation of the circle

#(7-a)^2+(3-b)^2=r^2# #(5-a)^2+(1-b)^2=r^2#

so #(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2# Developing both sides #49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2# simplifying #58-14a-6b=26-10a-2b# #4a+4b=32# #a+b=8# this is equation 1 Putting #(a,b)# in the equation of the line #b=7/9a+5# this is equation 2 Solving for a and b we obtain #(27/16,101/16)# as the center of the circle The radius #r=sqrt (85^2+63^2)/16=6.26# The equation of the circle is #(x-27/16)^2+(y-101/16)^2=39.2#

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Answer 2

Evelyn Arboleda

The equation of the circle is (x - 7)(x - 5) + (y - 3)(y - 1) = 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.