A circle has a center that falls on the line y = 7/4x +4 y=74x+4 and passes through ( 3 ,7 )(3,7) and (7 ,1 )(7,1). What is the equation of the circle?

1 Answer
Narad T.
Nov 1, 2016

The equation of the circle is (x+40/13)^2+(y+18/13)^2=10.51^2(x+4013)2+(y+1813)2=10.512

Explanation:

Let (a,b)(a,b) be the center of the circle
As the center lies on the line, we have b=(7a)/4+4b=7a4+4
Then (x-a)^2+(y-b)^2=r^2(x−a)2+(y−b)2=r2, and rr is the radius of the circle
So, (3-a)^2+(7-b)^2=r^2(3−a)2+(7−b)2=r2
and (7-a)^2+(1-b)^2=r^2(7−a)2+(1−b)2=r2
:. (3-a)^2+(7-b)^2=(7-a)^2+(1-b)^2
9-6a+a^2+49-14b+b^2=49-14a+a^2+1-2b+b^2
58-6a-14b=50-14a-2b
12b=8a+8=>3b=2a+2
Solving for (a,b) with the two equations
We find a=-40/13 and b=-18/13

Then r=sqrt((7+(40/13))^2+(1+(18/13))^2)
r=10.51
Having r, a, and b, we write the equation of the circle

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