In addition, this straight line passes through points P(- 1, 6) and ∴-A+6B+C=0 (1).

It is known that the center of the circle is (-3,2) and the radius of the circle is 2. ..

The circle is tangent to the straight line L, and the distance from the center of the circle to the straight line L is d=R=2. The distance formula from a point to a straight line is:

|-3A+2B+C|/√(A^2+B^2)=d=R=2.

|-3A+2B+C|^2=4(A^2+B^2) (2)。

Substitute the value of c from: C=A-6B of (1) to: (-2a-4b) 2 = 4 (a 2+b 2), 4 (a-2b) 2 = 4 (a 2+b 2).

(A-2B)^2=A^2+B^2.

A^2-4AB+4B^2=A^2+B^2.

3B^2-4AB=0，B(3B-4A)=0 ∵B≠0,∴3B-4A=0，A=3B/4。

C=A-6B=3B/4-6B=-2 1B/4。 Substitute the values of a, b and c into the linear equation:

(3B/4)x+By+(-2 1B/4)=0。

3x/4+y-2 1/4=0。

∴ 3x+4y-2 1 = 0。 -It's a linear equation.

2. The joint solution of x 2+y 2-x+y 2 = 0 and x 2+y 2 = 5 is: x=y+3. This is the intersection of two known circles. Because I'm not sure And x ∈ R. Therefore, let x= 1, then y=-2, and get an intersection point (1,-2); Let x = 3 and y = 0 intersect (3,0).

Because finding a circle passes through these two intersections and is substituted into the equation for finding a circle, we get:

( 1-a)^2+(-2-b)^2=r^2( 1)

(3-a)^2+(0-b)^2=r^2②。

( 1)-(2):( 1-a)2+(2+b)2-(3-a)2-B2 = 0。

a+b= 1 (3)。

Because the center (a, b) is on the straight line 3x+4y- 1=0, we get:

3a+4b= 1 (4)

Combine (3) and (4):

A=3，b=-2。 Substituting the values of a and b into (2) gives R=2.

An equation for finding a circle is: (x-3) 2+(y+2) 2 = 4.

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