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

1 Answer

(x-19)^2+(y-40/3)^2=2665/9

Explanation:

From the given two points (3, 7) and (7, 1) we will be able to establish equations

(x-h)^2+(y-k)^2=r^2

(3-h)^2+(7-k)^2=r^2" "first equation using (3, 7)

and

(x-h)^2+(y-k)^2=r^2

(7-h)^2+(1-k)^2=r^2" "second equation using (7, 1)

But r^2=r^2
therefore we can equate first and second equations

(3-h)^2+(7-k)^2=(7-h)^2+(1-k)^2

and this will be simplified to
h-3k=-2" "third equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The center (h, k) passes thru the line y=1/3x+7 so we can have an equation

k=1/3h+7 because the center is one of its points

Using this equation and the third equation,

h-3k=-2" "
k=1/3h+7

The center (h, k)=(19, 40/3) by simultaneous solution.

We can use the equation
(3-h)^2+(7-k)^2=r^2" "first equation
to solve for the radius r

r^2=2665/9

and the equation of the circle is

(x-19)^2+(y-40/3)^2=2665/9

Kindly see the graph to verify the equation of the circle (x-19)^2+(y-40/3)^2=2665/9 colored red, with points (3, 7) colored green, and (7, 1) colored blue, and the line y=1/3x+7 colored orange which contains the center (19, 40/3) colored black.

Desmos.comDesmos.com

God bless....I hope the explanation is useful.