4. Solution: Set the equation from the problem: x 2+y 2+6x-4+λ (x 2+y 2+6y-28) = 0.

That is, (1+λ) x2+(1+λ) y2+6x+6λ y-28λ-4 = 0.

Center (-3/( 1+λ), -3λ/( 1+λ)), ∴-3/(1+λ)+3 λ/(1+λ)-4 = 0.

The solution is λ=-7.

The equation for finding a circle is -6x 2-6y 2+6x-42y+28 * 7-4 = 0.

That is, x 2+y 2-x+7y-32 = 0.

5. The equation of a circle is (X- 1) 2+(Y-2) 2 = 5.

Center C( 1, 2)r= root number 5.

|CE|=|3-2-6|/ root number (1+9)= root number 10/2.

|AE|= root number (5- 10/4)= root number 10/2.

∴|AB|=2|AE|= root number 10

6. Let the radius r of the center (a, b) be the equation (x-a) 2+(y-b) 2 = r 2.

b=3a

|b|=r

(|a-b|/ radical number 2) 2 = r 2-7

The solution is a = 1

When a= 1, b=3, r=3.

When a=- 1, b=-3, r=3.

So the equation of a circle is (x+ 1) 2+(y+3) 2 = 9 or (x- 1) 2+(y-3) 2 = 9.

7. The radius of the center c: x 2+y 2-x+y = 0 (1/2,-1) is 5/2 of the radical sign.

Let the point where the center of the circle C( 1/2,-1) is symmetrical about the straight line x-y+ 1=0 be c "(x0, y0).

{(y0+ 1)/(0- 1/2)=- 1

(x0+ 1/2)/2

-(y0- 1)/2=0

X0=-2，y0=3/2。

So c "(-2,3/2)

Equation of circle (x+2) 2+(y-3/2) 2 = 5/4.