A circle has a center that falls on the line #y = 2/3x +7 # and passes through #(4 ,7 )# and #(1 ,2 )#. What is the equation of the circle?

1 Answer
Narad T.
Nov 9, 2016

The equation of the circle is #(x+15/19)^2+(y-123/19)^2=23.2#

Explanation:

Let the center of the circle be #(a,b)# and radius #r#
Then #b=2/3a+7#
The equation of the circle is #(x-a)^2+(y-b)^2=r^2#
Substituting the values of the 2 points
#(4-a)^2+(7-b)^2=r^2#
and #(1-a)^2+(2-b)^2=r^2#
#:.(4-a)^2+(7-b)^2=(1-a)^2+(2-b)^2#
#16-8a+a^2+49-14b+b^2=1-2a+a^2+4-4b+b^2#

#65-8a-14b=5-2a-4b#
#6a+10b=60##=>##3a+5b=30#
and the first equation is #3b-2a=21#
Solving for #a# and #b#, we get #(-15/19,123/19)# as the center of the circle
we calculate the radius of the circle
#r^2=(34/19)^2+(85/19)^2=23.21#
#r=4.82#

Equation of the circle
#(x+15/19)^2+(y-123/19)^2=23.2#

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