The center of the circle is on the perpendicular of the straight line AB. The slope of straight line ab Kab=(5- 1)/(6-0)=2/3, the slope of the vertical line in straight line AB is -3/2, and the midpoint of easy to find AB is (3,3). ∴ab's perpendicular equation is 3x+2y- 15=0, that is, the center of the circle is 3x+2y- 15 = 0, and the center of the circle is 3X+ 10Y+9=0, so the center of the circle is at the intersection of these two lines, and the center of the circle is (7.

(2) Let the circle equation be x? +y？ +Ax+By+C=0

Replace p and q:

-2A+4B+C=-20，3A-B+C=- 10

C=B- 10-3A

That is 5b-5a =-10 = = >; A-b = 2...( 1), let y=0, and get x? +Ax+C=0

So x 1+x2 =-a, x 1x2 = c, | x 1-x2 | = √ [(x 1+x2)? -4x 1x2]=√(A？ -4C)=6

Answer? -4C=36，A？ -4(B- 10-3A)=36，A？ -4B+ 12A+4=0，A？ +8A+ 12=0，(A+6)(A+2)=0

A=-6 or A =-2 = = =>；; B=-8 or-4 = = > C = 0 or -8.

So the circular equation is x? +y？ -6x-8y=0 or x? +y？ -2x-4y-8=0

2.( 1) The known point P(x, y) is any point on the circle (x+2) 2+y 2 =1,and the straight line is 3x+4y+ 12=0.

Then, the distance from the center O (-2,0) to the straight line D = | 3 * O(-2)+4 * 0+12 |/5 = 6/5.

So, straight lines and circles are separated.

Then, the intersection circle O is the perpendicular of the straight line, and the intersection circle is at p1; The intersecting circle of the opposite extension lines of the vertical line is at P2.

Then: P 1 distance to the straight line =(6/5)-r= 1/5 is the minimum value;

The distance from P2 to the straight line =(6/5)+r= 1 1/5 is the maximum.

2. solution: let z=x-2y.

Then z takes the maximum value when the straight line y=(x-z)/2 is tangent to the circle.

The simultaneous equation of a line and a circle can give 5x2+(16-2z) x+(z2+12) = 0.

Discriminant△ = 0

So you can get z =-2 √ 5.

So max=-2+√5, min=-2-√5.

3. Solution: (y-2)/(x- 1) represents the slope from a point on the circle to (1, 2).

Suppose the slope is k and the linear equation is y-2=k(x- 1), that is, kx-y+(2-k)=0.

Find the critical value when a straight line is tangent to a circle.

So |-2k+0+2-k |/√ (k2+1) =1.

So k = (3 √ 3)/4.