Equation of a circle whose center is on the straight line x-y-4=0 and passes through the intersection of two circles x*x+y*y-4x-6=0 and x*x+y*y-4y-6=0.

X*x+y*y-4x-6=0，x*x+y*y-4y-6=0。

The solution is x 1 = 3, y1= 3; x2=- 1，y2=- 1

So the intersection of two circles is m (3,3) and N(- 1,-1).

Because the circle passes through these two points and connects MN, MN is a chord of the circle.

Now ask the perpendicular line in MN,

Because the slope of MN k1= (3+1)/(3+1) =1.

Therefore, its vertical slope k2 =-1; The midpoint p coordinate of MN is (1, 1).

So the median vertical line is y =-x+2.

The middle vertical line and the topic give the intersection point of the straight line, that is, the center of the circle, and simultaneously establish two equations.

y=-x+2

x-y-4=0

X = 3, y =- 1, so the coordinate of the center o is (3,-1).

The line OM is the radius of the circle.

R = Under the radical sign (0+4 2) = 4

So the equation of a circle is (x-3) 2+(y+ 1) 2 = 16.