Question 246624
Let's say the center of the circle is C. Draw pependicular lines from C to the two tangent lines created by y = |x|. Let's say the two lines intersect the tangent line at points A and B respectively. Let's say the Origin is point O with coordinates (0,0).

The figure CAOB is a square with sides of length 6. Triangle CAO is a right isosceles triangle with sides of length 6. So the hypotenuse CO of CAO is sqrt(36+36) = sqrt(72) = 6*sqrt(2). 

The cooredinates of C then are (0,6*sqrt(2)). 

So we have a circle with radius 6 and center C at (0,6sqrt(2).

Plug these values into the equation of the standard form of a circle:

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius of the circle.