A circle has a center that falls on the line y=13x+7 and passes through (3,7) and (7,1). What is the equation of the circle?

1 Answer
Leland Adriano Alejandro
Jul 25, 2016

(x−19)2+(y−403)2=26659

Explanation:

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

(x−h)2+(y−k)2=r2

(3−h)2+(7−k)2=r2 first equation using (3,7)

and

(x−h)2+(y−k)2=r2

(7−h)2+(1−k)2=r2 second equation using (7,1)

But r2=r2
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=13x+7 so we can have an equation

k=13h+7 because the center is one of its points

Using this equation and the third equation,

h−3k=−2
k=13h+7

The center (h,k)=(19,403) by simultaneous solution.

We can use the equation
(3−h)2+(7−k)2=r2 first equation
to solve for the radius r

r2=26659

and the equation of the circle is

(x−19)2+(y−403)2=26659

Kindly see the graph to verify the equation of the circle (x−19)2+(y−403)2=26659 colored red, with points (3,7) colored green, and (7,1) colored blue, and the line y=13x+7 colored orange which contains the center (19,403) colored black.

![Desmos.com](useruploads.socratic.org)

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

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