A circle has a center that falls on the line y = 12/7x +3 and passes through ( 9 ,5 ) and (8 ,7 ). What is the equation of the circle?

1 Answer
Narad T.
Oct 23, 2016

The center of the circle is (-7/11,91/44)
And the radius is 9.95
The equation of the circle is (x+7/11)^2+(y-91/44)^2=98.9

Explanation:

Let (a,b) be the center of the circle and r the radius
The equation of a circle is (x-a)^2+(y-b)^2=r^2
So we apply this to the 2 points and we get

(9-a)^2+(5-b)^2=r^2
(8-a)^2+(7-b)^2=r^2
So we can combine these 2 equations
(9-a)^2+(5-b)^2=(8-a)^2+(7-b)^2
Developing

81-18a+a^2+25-10b+b^2=64-16a+a^2+49-14b+b^2

Simplifying
81-18a+25-10b=64+49-16a-14b
106-113=18a-16a-14b+10b
-7=-2a-4b => 4b=-2a+7 this equation 1
From the equation of the line, we get
y=(12x)/7+3 => b=(12a)/7+3 this is equation 2

So-a/4+7/4=(12a)/7+3=> a=-7/11
and 4b=-2*-7/11+7=> b=91/44

from these we can calculate r
r=sqrt((8--7/11)^2+(7-91/44)^2)=9.95

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