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 ,5 )(7,5). What is the equation of the circle?

1 Answer
Narad T.
Oct 26, 2016

The equation of the circle is (x-32)^2+(y-60)^2=3650(x−32)2+(y−60)2=3650

Explanation:

Let the center of the circle be (a,b)(a,b) and radius r
Then the equation of the circle is
(x-a)^2+(y-b)^2=r^2(x−a)2+(y−b)2=r2

Then, (3-a)^2+(7-b)^2=r^2(3−a)2+(7−b)2=r2
and (7-a)^2+(5-b)^2=r^2(7−a)2+(5−b)2=r2

Equqting the equations
(3-a)^2+(7-b)^2=(7-a)^2+(5-b)^2(3−a)2+(7−b)2=(7−a)2+(5−b)2
developing,
9-6a+a^2+49-14b+b^2=49-14a+a^2+25-10b+b^29−6a+a2+49−14b+b2=49−14a+a2+25−10b+b2

9-6a-14b=25-14a-10b9−6a−14b=25−14a−10b
8a-4b=168a−4b=16
2a-b=42a−b=4 this is equation 1
Also, y=(7x)/4+4y=7x4+4
So b=(7a)/4+4b=7a4+4
Solving for a and be, we get a=32a=32 and b=60b=60
We calculate the radius r^2=(7-32)^2+(5-60)^2r2=(7−32)2+(5−60)2

r^2=25^2+55^2r2=252+552=>⇒ r=60.41r=60.41

Equation of circle is
(x-32)^2+(y-60)^2=3650(x−32)2+(y−60)2=3650

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