A circle has a center that falls on the line y = 7/9x +7 y=79x+7 and passes through ( 4 ,1 )(4,1) and (3 ,7 )(3,7). What is the equation of the circle?

1 Answer
Roy E.
Dec 31, 2016

The circle is
(x+129/22)^2+(y-161/66)^2=216413/2178(x+12922)2+(y−16166)2=2164132178

Explanation:

The general equation of the circle, center (a,b)(a,b), radius rr is
(x-a)^2+(y-b)^2=r^2(x−a)2+(y−b)2=r2.

Since the center lies on a given line, b=7/9a+7b=79a+7.
Since two given points line on the circle:
(4-a)^2+(1-b)^2=r^2=(3-a)^2+(7-b)^2=r^2(4−a)2+(1−b)2=r2=(3−a)2+(7−b)2=r2
Hence 16-8a+cancel(a^2)+1-2b+cancel(b^2)=9-6a+cancel(a^2)+49=14b+cancel(b^2)

So the problem reduces to solving for a, b the simultaneous equations above.

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