Find the equation of the circle touching the line x + y = 4 at (1,3) and having a radius of sqrt. of 2 solve in two ways.

Answer 1

The circle has equation
(x-h)^2 + (y-k)^2 = 2
A radius with slope 1 perpendicular to the line (which has slope -1) is thus
y=x+2

So, the circle has equation

(x-h)^2 + (y-(h+2))^2 = 2
Since it contains (1,3), we have
(1-h)^2 + (3-(h+2))^2 = 2
h=0,2
and there are two circles that touch the line:
x^2 + (y-2)^2 = 2
(x-2)^2 + (y-4)^2 = 2

see the graphs at

http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+(y-2)%5E2+%3D+2,+(x-2)%5E2+%2B+(y-4)%5E2+%3D+2,+x%2By%3D4

Another way to look at it is this:

Since the radius is √2, and lies on the line y=x+2, the center lies 1 unit to the left or right of (1,3). So, the center has x-coordinate 0 or 2. The rest follows.