Mathematics problems about circle It's a little harder

Mathematics problems about circle It's a little harder

The 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 and X * x + y * y-4y-6 = 0
The results show that X1 = 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 find the vertical bisector of Mn,
Because the slope of Mn K1 = (3 + 1) / (3 + 1) = 1
So the slope of the vertical bisector K2 = - 1; the P coordinate of the middle point of Mn is (1,1)
So the vertical bisector is y = - x + 2
The intersection of the bisector and the subject line is the center of the circle, and two equations are established simultaneously
y＝－x＋2
x-y-4=0
The solution is x = 3, y = - 1, so the coordinates of the center o point are (3, - 1)
Connecting OM is the radius of the circle
R = under the root sign (0 + 4 ^ 2) = 4
So the equation of the circle is (x-3) ^ 2 + (y + 1) ^ 2 = 16